Download Physics Newton`s Laws Worksheet Solutions

Physics
Newton’s Laws Worksheet Solutions
Part I
1. A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off so that friction
between the cart and the track now acts. The cart comes to rest after traveling 1 m. The
experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How
far does the cart travel before coming to rest?
(a) 1 m
(b) 2 m
(c) 3 m
(d) 4 m
1
2
Answer: (d) 2 mv = f ∆x so if velocity is doubled then distance traveled will be four times.
2. A constant force is exerted on a cart that is initially at rest on a track. Friction between the
cart and the track is negligible. The force acts for a short time interval and gives the cart a
certain final speed. To reach the same final speed with a force that is only half as big, the
force must be exerted on the cart for a time interval
(a) four times as long as for the stronger force
(b) twice as long as for the stronger force
(c) half as long as for the stronger force
(d) a quarter as long as for the stronger force
Answer (b): If the force is half as big but the mass is the same then the acceleration is half
as big so it takes twice as long to reach the same velocity.
3. A constant force is exerted for a short time interval on a cart that is initially at rest on
frictionless track. This force gives the cart a certain final speed. The same force is exerted
for the same length of time on another cart, also initially at rest, that has twice the mass of
the first one. The final speed of the heavier cart is
(a) one-fourth that of the lighter cart
(b) half that of the lighter cart
(c) the same as that of the lighter cart
(d) double that of the lighter cart
Answer (b): If the force is the same but the mass is double then the acceleration is half. So
if the time interval is kept the same than the change in velocity is half.
4. A rocket of mass 10000 kg on the surface of the earth accelerates upward at a rate of 4 m/s2 .
The force provided by the rocket engines must be
(a) 40000 N
(b) 100000 N
(c) 140000 N
(d) 160000 N
Answer (c): Call the upward force T for thrust. Fnet = T − W = ma ⇒ T = W + ma =
mg + ma ≈ 140000 N.
Part II
1. A block pushed along the floor with velocity ⃗v0 slides a distance d after the pushing force is
removed.
(a) If the mass of the block is doubled but the initial velocity is unchanged what is the
distance the block slides before stopping?
The distance the block travels depends on the size of the acceleration that is slowing it
down. Doubling the mass doubles the inertia, but it also doubles the friction and so the
acceleration is unchanged. Thus the stopping distance is unchanged in this model.
(b) If the initial velocity is doubled to 2⃗v0 but the mass is not changed, what is the distance
the block slides?
An object travelling twice the speed but the same acceleration will take twice as long to
slow down. Since its average velocity is also twice as much it will travel four times the
distance before stopping.
2. Suppose you press a book against the wall with your hand. The book is not moving. Draw
a free-body diagram identify all the forces on the book. What happens to each of the forces
you identified if the force you push with is decreased but the book doesn’t slip? What if it
slips? Which forces decrease, which stay the same, and which increase?
The pushing force on the book is equal and opposite to the normal force of the wall on the
book. The static friction up balances the weight acting down. When the book is pushed less
hard the normal force also decreases. The static friction and the weight stay the same. Note:
the maximum static friction decreases because the normal force decreases, but as long as the
friction required to balance the weight is less than the maximum friction then the book will
not slide. The book slides when the maximum friction is less than the weight.
3. Three forces cause a 1.0 kg object to accelerate. Two of the forces are shown in the free-body
diagrams below. For each, find the components of the missing force given the acceleration
shown. Draw the third force on your diagram.
4. A block of weight W = 20 N is pushed with
a force F = 30 N using a stick which is at
an angle of θ = 37o above the horizontal as
shown. The coefficient of kinetic friction
between the table and the block is µ =
0.25.
θ
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(a) Draw a free body diagram showing all the forces acting on the block.
N
θ
f
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F
W
(b) Calculate the value of the normal force between the block and the table.
The normal force balances the weight W and the vertical component of the downward
force Fy . N = W + Fy = 20 + 30 sin 37 = 38.1 N
(c) Find the acceleration of the block.
The acceleration is to the right and a = Fnet /m were m = W/g = 20/9.8 = 2.04 kg and
Fnet = Fx − f . Now friction is f = µN = 0.25 × 38.1 = 9.5 N and Fx = 30 cos (37) =
24.0 N so Fnet = 24.0 − 9.5 = 14.5 N. So a = 14.5/2.04 = 7.1 m/s2
(d) The force F is decreased until the acceleration is zero. Find this value of F .
If acceleration is zero then Fnet = 0 which means friction exactly balances the horizontal
component of F . We treat F as a variable and express normal force N in terms of F .
So N = 20 + F sin 37 and hence friction is f = 0.25(20 + F sin 37) = 5 + 0.25F sin 37.
Then Fx = F cos 37 and so for equilibrium F cos 37 = 5 + 0.25F sin 37 Solving for F we
find F cos 37 − 0.25F sin 37 = 5 so F = 5/(cos 37 − 0.25 sin 37) = 7.71N
(e) There is a critical value of the angle θ for which no force is able to accelerate the block.
Find this value of θ.
If we inspect the last equation we see that if the denominator were zero then the expression for F would be undefined. Thus, replacing 37o with the a variable angle θ and
setting the denominator equal to zero we find a critical angle, above which no force
could cause the block to accelerate. We must solve the equation cos θ − 0.25 sin θ = 0
or cos θ = 0.25 sin θ. Now dividing by cos θ and noting that sin θ/ cos θ = tan θ we have
1 = 0.25 tan θ ⇒ tan θ = 1/0.25 = 4 ⇒ θ = tan−1 4 = 76o So the critical angle is 76o .