Question

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

Answer 1

Answer:

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

Answer 2

Answer:

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