In order to move something against a constant frictional force, you must exert a force greater than or equal to the frictional force. So you exert at least F = 230 N in order to move the crate directed along the displacement direction. Therefore:

W = Fd cosq = (230 N)(4.0 m)(1) =

(b) 4.0-m vertically?
__Solution:__

This one is similar to the firefighter problem. The force you must exert
must be at least equal to the weight of the crate.

W = 1200 N (4.0 m) = **4800 J (answer)**

The reference point for this part is A, therefore our y = 0 is at point A

Point B is a distance y

U

Point C is a distance y

U

(b) What is the change in potential energy
as it goes from B to C?
__Solution__:**
**DU
= U

(c) What is the potential energy at B and
C relative to point C?
__Solution__:

The reference point for this part is C, therefore our y = 0 is at point
V

Point B is a distance y_{b} = -(25 - 10)m = -15 m below our reference
point

U_{b} = mgy_{b} = (1000 kg)(9.80 m/s^{2})(-15 m)
= **-147 kJ**

Point C is a distance y_{c} = 0

U_{c} = mgy_{c} = (1000 kg)(9.80 m/s^{2})(0) =
0

(d) What is the change in potential energy
as it goes from B to C?
__Solution__:**
**DU
= U

U

The initial energy of the system is purely elastic potential energy since the spring is compressed and the mass is initially at rest. Therefore:

E

The final energy of the system is purely kinetic since the mass is now moving and the spring is now back at its equilibrium position. Therefore:

E

By conservation of energy: E

½(0.100 kg)v

v

The energy at A is purely gravitational potential energy since the car starts from rest. Therefore:

E

The energy at the bottom of the hill is purely kinetic, no potential energy since y = 0. Therefore

E

Using conservation of energy, we can easily solve for v:

½Mv

v

The energy at B is a combination of potential and kinetic energy, therefore:

E

E

½ Mv

v

The change in energy is equal to the work done by the non-conservative force:

DE = E

W

d = (E

f

The initial energy is purely kinetic and the final energy is 0:

E

d = (0 - ½mv

= (2.0)/(0.98)

(a) Find the tension on the cable

The system is NOT in equilibrium since there is a net acceleration. Therefore:

T - mg = ma

T = m (g + a) = (220 kg)( 9.80 + 1.47)

(b) Find the net work done on the crate
__Solution__:

The net work done is done by the net force: F_{net} = ma = (220
kg)(1.47 m/s^{2}) = 323 N

W_{net} = (323 N)(21.0 m) = **6,800 J**

(c) the work done by gravity on the crate
__Solution__:

W_{g} = (F_{g} cos 180)(21.0 m) =** -45,300 J**

(d) Find the work done by the cable on the
crate
__Solution__:**
**W

(e) find the final speed of the crate assuming
it started from rest.
__Solution__:

The werk-energy theorem states that the change in the kinetic energy of
the system is equal to the net wok done. The system starts from rest,
therefore the final kinetic energy = 0.

DK = 1/2(220
kg)v_{f}^{2}

1/2(220 kg)v_{f}^{2}= 6800 J

v_{f}^{2}= 2(6800 J)/(220 kg) = 61.8

**
v _{f}= 7,86 m/s**

First, find the work needed to lift the piano: W = mgh = (285 kg)(9.80 m/s

P = W/t

t = W/P = (44,688 J)/(1750 W) =

K = ½ mv

P = W/t = 1000 J/(1.0 s) =