Physics 1101: Newton's Laws (Review) Solutions

1. In the diagrams below, a ball is on a flat horizontal surface. The velocity and external forces acting on the ball are indicated. Describe qualitatively how motion the motion of the ball will change.

First determine the direction of the net force on the ball. From Newton's Second Law, F = ma, so the direction of the net force and resulting acceleration are in the same direction. As time passes, the direction of the ball will tend to point in the direction of the acceleration.





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2. The 80.0 kg male partner of a figure skating duo pushes his 60.0 kg female partner with a force of 70.0 N. Find the acceleration of both partners.

According to Newton's Third Law, the woman must exert 70.0 N on the male skater but in the opposite direction, in other words, -70.0 N, where the minus sign indicates the direction opposite to the direction that the man pushes the woman.

To find the acceleration of the male skater we use Newton's Second Law,

a_man = F_{woman­on­man} / m_man = -70 N / 80.0 kg = -0.875 m/s².

The minus sign indicates that the man moves backwards which is what one would expect.

Similarly the acceleration of the female skater is

a_woman = F_{man­on­woman} / m_woman = +70 N / 60.0 kg = +1.17 m/s².

The plus sign indicates that the woman moves backwards (forward from the man's perspective) which is what one would expect.



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3. What is the normal force on the object of mass m₁ shown in the diagram? What is the acceleration of the object? In the diagram F₁ = 25 N, F₂ = 15 N, and m₁ =20 kg.

This problem deals with forces and acceleration, which suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). In the diagram we show the givens forces F₁ and F₂. As well, since the object has mass, it has weight. Since the object touches the table top, there is a normal force from the table top through the object. We assume that the object will accelerate to the right.



Next we apply Newton's Second Law separately to the i and j components:

i	j
Fx = max	Fy = may
F1cos(35) + F2cos(43) = m1a	N - m1g - F1sin(35) + F2sin(43) = 0

Thus the normal force is:

N = m₁g + F₁sin(35) - F₂sin(43) = (20)(9.81) +25sin(35) - 15sin(43) = 200.3 N .

The acceleration is:

a = [F₁cos(35) + F₂cos(43)]/m₁ = [25cos(35) + 15cos(43)]/20 = 1.572 m/s² .



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4. The apparent weight of a person in an elevator is 7/8ths of his actual weight. What is the acceleration (including the direction) of the elevator?

This problem deals with a force, weight, and acceleration. That suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). Since the person has mass, he has weight. Since the person touches the floor of the elevator, there is a normal force from the floor. We will assume that the person will accelerate upwards.



Applying Newton's Second Law yields,

F_y = ma_y

N - mg = ma.

This is not enough information to solve the problem, since we have one equation but two unknowns, N and a. However, the apparent weight is just N, so the first sentence of the problem is N = (7/8)mg. Now we can find a,

a = (N - mg) / m = [(7/8)mg - mg]/m = -g/8 = -1.23 m/s² .

The minus sign indicates that, contrary to our initial assumption, the elevator accelerates downwards at 1.23 m/s².



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5. A person is standing on a weigh scale in an elevator. When the elevator is accelerating upward with constant acceleration a, the scale reads 867.0 N. When the elevator is accelerating downwards with the same constant acceleration a, the scale reads 604.5 N. Determine the magnitude of the acceleration a, the weight of the person, and the mass of the person.

This problem deals with a force, the scale reading, and acceleration. That suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). Since the person has mass, he has weight. Since the person touches the scale, there is a normal force from the scale. The scale reading is a measure of this normal force. Since there are two cases, we draw an FBD for each case.



up	down
Fy = may	Fy = may
N2 - mg = +ma	N2 - mg = -ma

So we have two equation in two unknowns, m and a. First we add the two equations together to get N₁ - mg + N₂ - mg = 0. This becomes m = (N₁ + N₂) / 2g = 75.0 kg.

Using the first equation, we get a = (N₂ - mg) / m = 1.75 m/s².



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6. A car is traveling over the crest of a small semi-circular hill of radius R = 750 m. How fast would it have to be traveling for it to leave the ground?

Notice that the car is traveling in part of a circle, that suggests a centripetal acceleration. For the car to leave the ground, the normal force between the car and the ground must be zero. Since this problem deals with a force, the normal, and acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for the car at the crest of the hill. The car has mass, so it has weight. There is the normal force. By definition, the centripetal acceleration points towards the centre of the hill; in this case down.



Applying Newton's Second Law, F_y = ma_y, we get N - mg = -mv²/R. The car loses contact when N = 0, so our equation for the speed becomes v = [gR]^½ = [9.81750]^½ = 85.8 m/s = 309 km/h.



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7. In the diagram below, block A weighs 100N. The coefficient of static friction between the block and the surface on which it rests is 0.40. Find the maximum weight w for which the system will remain in equilibrium.






A system in equilibrium does not accelerate, so a = 0. Since this problem deals with forces, friction and weight, and we known that the acceleration is zero, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each block and the knot. We need a FBD for the knot since this is where the tensions all meet. A knot is massless since we are dealing with massless ropes. Each block has mass, so each has weight. Each rope represents a different tension. Block A is on a table so there is a normal force from the table through A. If there were no friction, Block A would move forward. We deduce that the maximum static friction points left.



Block A		knot		w
i	j	i	j	j
Fx = max	Fy = may	Fx = max	Fy = may	Fy = may
T1 - fs MAX = 0	N - mg = 0	T2cos(40) - T1 = 0	T2sin(40) - T3 = 0	T3 - w = 0

Besides our equations in the last row above, we also have the definition f_{s MAX} = μN. Our equation in the second column tells us that N = mg = 100 N, so we know f_{s MAX} = μN = 0.40100 N = 40 N. The first equation tells us that T₁ = f_{s MAX} = 40 N. The third equation yields, T₂ = T₁/ cos(40) = 52.2 N. The fourth equation yields T₃ = T₂sin(40) = 33.6 N. The equation in the last column indicates that w = T₃ = 33.6 N. So the maximum weight of the hanging block is 33.6 N.



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8. A force F pushes on a 25-kg box as shown in the figure below. The coefficient of static friction between box and incline is μ_s = 0.20 . Find the range of values for which the block remains stationary.






Since this problem deals with forces, weight and friction, and we known that the acceleration is zero since the block not moving, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each case. Since the block has mass, it has weight. It is on the incline, so there is a normal. For a stationary object, we must be dealing with static friction. As the problem asks for a range of values, we must examine the cases where friction acts up the incline and down the incline.



down
i	j
Fx = max	Fy = may
F1cos(τ) - fs MAX - mgsin(θ) = 0	N - mgcos(θ) - F1sin(θ) = 0

Since we know f_{s MAX} = μ_sN, and the second equation says that N = mgcos(θ) + F₁sin(θ), we have f_{s MAX} = μ_s[mgcos(q ) + F₁sin(θ)]. Substituting this into the first equation yields,

F₁cos(θ) - μ_s [mgcos(θ) + F₁sin(θ)] - mgsin(θ) = 0 .

Collecting similar terms we get

F₁[cos(θ) - μ_ssin(θ)] - mg[sin(θ) + μ_scos(θ)] = 0 .

So we find that

F₁ = mg[sin(θ) + μ_scos(θ)] / [cos(θ) - μ_ssin(θ)] = 306 N.

up
i	j
Fx = max	Fy = may
F2cos(θ) + fs MAX - mgsin(θ) = 0	N - mgcos(θ) - F2sin(θ) = 0

Since we know f_{s MAX} = μ_sN, and the second equation says that N = mgcos(θ) + F₂sin(θ), we have f_{s MAX} = μ_s [mgcos(θ) - F₂sin(θ)]. Substituting this into the first equation yields,

F₂cos(θ) - μ_s [mgcos(θ) - F₂sin(θ)] - mgsin(θ) = 0 .

Collecting similar terms we get

F₂[cos(θ) + μ_s sin(θ)] - mg[sin(θ) - μ_s cos(θ)] = 0 .

So we find that

F₁ = mg[sin(θ) - μ_scos(θ)] / [cos(θ) + μ_s sin(θ)] = 134 N.

So when the applied force, F, is between 134 N and 306 N, the block remains stationary.



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9. What must be the minimum nacceleration of the cart pictured below in order that block A does not fall? The coefficients of friction between the block and the cart are μ_s and μ_k/sub>.






Since this problem deals with a force, friction, and an acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each object. The cart and the block must have the same acceleration. As we will see, we only need an FBD for the block to solve the problem. We would need another FBD if the question asked what force was accelerating the cart. The block has mass, so it has weight. It is touching the cart, so there is a normal. We have to decide which way the static friction points. If the block is not to fall, it must be upwards.






i	j
Fx = max	Fy = may
N = ma	fs MAX - mg = 0

We know that f_{s MAX} = μ_sN. The first equation gives us N, so f_{s MAX} = μ_sma. The second equation says, f_{s MAX} = mg. So therefore μ_sma = mg or a = g / μ_s.



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10. Three blocks with masses 6 kg, 9 kg, and 10 kg are connected as shown below. The coefficient of kinetic friction between the 10 kg block and the table is 0.2 . Find the acceleration of the system. Find the tension in each cord.






Since this problem deals with forces, friction and tension, and acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each object. The three blocks must have the same acceleration, though in different directions. Each block has mass, so each has weight. The block on the table has a normal acting on it. The tension is the same throughout each rope as long as we deal with ideal massless, frictionless pulleys. Presumably, the 9-kg block will move downwards, therefore kinetic friction on the middle block points left.



6-kg block	10-kg block		9-kg-block
j	i	j	j
Fy = may	Fx = max	Fy = may	Fy = may
T1 - m1g = m1a	T2 - T1 - fk = m2a	N - m2g = 0	T2 - m3g = -m3a

We know that f_k = μ_kN. The third equation gives us N = m₂g, so f_k = μ_km₂g. The first equation gives us an expression for T₁, T₁ = m₁g + m₁a. The fourth equation gives an expression for T₂, T₂ = m₃g - m₃a. With these results, we can rewrite the second equation as

[m₃g - m₃a] - [m₁g + m₁a] - μ_km₂g = m₂a .

Collecting the terms involving a, and rearranging yields

a = g [m₃ - m₁ - μ_km₂] / [m₃ + m₂ + m₁] .

Substituting in the given values, one finds a = 0.3924 m/s².

Using this value for a, T₁ = m₁g + m₁a = 61.2 N. As well, T₂ = m₃g - m₃a = 84.8 N.



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11. A 10-kg block is hanging from a spring with spring constant K = 1000 N/m. The spring is attached to the ceiling of an elevator. The elevator is currently moving upwards at 10 m/s and slowing down at 1.0 m/s². How much is the spring stretched?

The elevator is moving upwards but slowing so the acceleration points downwards.



After drawing the free-body diagram, we apply Newton's Second Law and get kx – mg = –ma. Solving for x we find

x = m(g – a)/K = 10(9.81 – 1.0)/1000 = 0.088 m .

The spring is stretched 8.8 cm.

Questions?mike.coombes@kwantlen.ca