A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of - 13.0 N acts on the cart and its velocity becomes 3.60 m/s.

a) What is the change in kinetic energy of the cart? - 369.36

b) How much work was done on the cart?

c) How far did the cart move while the force acted? m

a. - 369.36J

b. - 369.36J

c. 28.41m

Explanation:

a. ΔK. E = 1/2 * m * (v₂²-v₁²)

ΔK. E = change in kinetic energy

v₂ = final velocity = 3.6m/s

v₁ = initial velocity = 7.2m/s

m=mass=19kg

ΔK. E = 1/2 * 19 * (3.6²-7.2²)

=1/2*19 * (12.96-51.84)

=9.5 * (-38.88)

= - 369.36J

b. the workdone is equal to the change in kinetic energy

∴ W. D = - 369.36J

c. W. D = force * distance

distance = W. D:force

= - 369.36 : - 13.0

=28.41m

a. - 369.36J

b. - 123.9J

c. 9.52m

Explanation:

From the expression for kinetic energy

K. E=1/2mv^2

Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is

K. E=1/2*19 (3.6²-7.2²)

K. E = - 369.36J

b. to determine the workdone by the force, we determine the distance moved.

But the acceleration is from

F=ma,

a=f/m

a=-13/19

0.68m/s²

the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

Hence the work done is

W=force * distance

W=-13*9.52

W=-123.9J

d. the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m