Question

To practice Problem-Solving Strategy 6.1 Work and Kinetic Energy. Your cat "Ms." (mass 8.00 kg ) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined 19.0 ∘ above the horizontal. Since the poor cat can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 60.0 N force parallel to the ramp. If Ms. is moving at 2.00 m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline?

Answer 1

given,

mass of the cat = 8 kg

length of the ramp = 2 m

inclined at = 19°

force exerted = 60 N

cat is moving at speed = 2 m/s

velocity when it reaches at the top = ?

using work energy theorem

work done by the gravity = - mgL sin θ

total work (w)

= F x - mgL sin θ

= 60 x 2 - 8 x 9.8 x 2 sin 19°

= 68.95 J

initial kinetic energy

`K E_i = (1)/(2)mv_i^2`

`K E_i = (1)/(2)* 8 * 2^2`

`K E_i = 16\ J`

change in kinetic energy is work done

`K_f - K_i = W`

`(1)/(2)mv^2 = W+k_i`

`v = \sqrt{(2(W+k_i))/(m)}`

`v = \sqrt{(2(68.95+16))/(8)}`

v = 4.61 m/s