Download Answers to the Problems on May the Force Be With You

Answers to May The Force Be With You
2. Calculate the acceleration of a 25-kg object under the influence of
a force of 250 N to the North.
Fnet = m a
m
m
a =
Fnet
m
=
250 N to the North
25 kg
a = 10 m/s2 to the North
3. Calculate the force required to increase the velocity of a car
(mass = 1.00 x 103 kg) from 4.0 m/s to 10.0 m/s in 6.0 s.
Fnet = m a
a =
a = ?
vf - vi
t
=
10.0 m/s - 4.0 m/s
6.0 s
a = 1.0 m/s2
F = ma
F = 1000 N
= ( 1000 kg )( 1.0 m/s2 )
6. A 20.0-kg box is pushed across the floor with a force of 150.0 N.
The force of friction Ff = -50.0 N.
(a) Calculate the net force Fnet.
(b) Calculate the acceleration of the box.
m = 20 kg
Ff = - 50 N
(a)
Fnet = FA + Ff
= ( + 150 N ) + ( - 50 N )
Fnet = + 100 N
(b)
Fnet = m a
m
m
a =
Fnet
m
a = 5.0 m/s2
=
100 N
20 kg
FA = 150 N
7. A car of mass 1.00x 103 kg is pushed across a level parking lot.
(a) Calculate the weight of the car.
(b) If the coefficient of friction  = 0.25, calculate the force of friction.
(a)
Wt. = m g = ( 1.00 x 103 kg )( 9.80 m/s2 )
Wt. = 9800 N
(b)
Ff = μ FN
FN = Wt. = 9800 N
= ( 0.25 )( 9800 N )
Ff = 2450 N
8. A force of 60.0 N is necessary to keep a cart moving across the floor
at constant velocity. The coefficient of friction  = 0.30.
(a) Find the force of friction.
FA = 60 N
(b) Calculate the mass of the cart.
Ff
(a) At constant velocity, FA = Ff
Ff = 60 N
(b)
Ff = μ FN = μ Wt.
Ff = μ m g
μg
μg
m =
Ff
μg
m = 20.4 kg
=
60 N
( 0.30 )( 9.8 m/s2 )
9. A woman that has a mass of 48.0 kg waits for an elevator.
(a) Calculate her weight in newtons and pounds.
Wt. = m g = ( 48.0 kg )( 9.80 m/s2 )
Wt. = 470 N
Conversion factor:
470.4 N
1 lb
4.45 N
1 lb = 4.45 N
= 106 lb
9. A woman that has a mass of 48.0 kg waits for an elevator.
(b) The doors open. In the elevator is a bathroom scale calibrated
in newtons. She steps on the scale. What does the scale read?
The elevator is not accelerating, so the force
on her is just due to gravity
Scale reads 470 N
9. A woman that has a mass of 48.0 kg waits for an elevator.
(c) The elevator accelerates upward at 1.82 m/s2.
What does the scale read now?
Elevator is accelerating her upwards ; elevator must exert a
force according to Newton’s 2nd Law:
Fnet = m a = ( 48.0 kg )( 1.82 m/s2 ) = 87.4 N
Elevator must also exert a force counteracting her weight;
so the total force exerted by the elevator (and read by the
scale) is
Ftotal = ( 470.4 N ) + ( 87.4 N )
Ftotal = 558 N
10. A Rocket has a mass of 1.50 x 103 kg. The engines exert a thrusting
force of 4.50 x 104 N. Calculate the acceleration of the rocket
(remember to take into account the weight).
FA = + 45 000 N
Wt. = m g
= ( 1500 kg )( 9.8 m/s2 )
= 14 700 N
a=?
m = 1500 kg
Wt.
10. A Rocket has a mass of 1.50 x 103 kg. The engines exert a thrusting
force of 4.50 x 104 N. Calculate the acceleration of the rocket
(remember to take into account the weight).
FA = + 45 000 N
Wt. = m g
Fnet
2
= ( 1500 kg )( 9.8 m/s )
a=?
= 14 700 N
m = 1500 kg
Fnet = FA + Wt.
= ( + 45 000 N ) + ( - 14 700 N )
Fnet = + 30 300 N
Wt. = 14 700 N
10. A Rocket has a mass of 1.50 x 103 kg. The engines exert a thrusting
force of 4.50 x 104 N. Calculate the acceleration of the rocket
(remember to take into account the weight).
FA = + 45 000 N
Fnet = m a
m
m
a =
Fnet
Fnet = + 30 300 N
a=?
m = 1500 kg
m
=
+ 30 300 N
1500 kg
a = + 20.2 m/s2
Wt. = 14 700 N