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.
2 answers:
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:
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
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