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

Answer

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

$K E_i = \dfrac{1}{2}mv_i^2$

$K E_i = \dfrac{1}{2}\times 8 \times 2^2$

$K E_i = 16\ J$

change in kinetic energy is work done

$K_f - K_i = W$

$\dfrac{1}{2}mv^2 = W+k_i$

$v = \sqrt{\dfrac{2(W+k_i)}{m}}$

$v = \sqrt{\dfrac{2(68.95+16)}{8}}$

v = 4.61 m/s