Your cat “Ms.”(mass 9.00 kg) is trying to make it to the top of a frictionless ramp 2.00m long and inclined 19.0 degrees 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 50.0N force parallel to the ramp. If Ms. is moving at 2.00m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline? What is the cat's speed v2 when she reaches the top of the incline?

Answer 1

force of gravity down ramp = m g sin19 = 9 * 9.81 * .3256 = 28.7 N
push force up ramp = 50 N
net force up = 50 - 28.7 = 21.26 N up ramp
work done by push = 21.26 * 2 = 42.5 Joules

that 42.5 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
42.5 = 57.5 +(1/2)9(V^2 - 2^2)
- 15 = 4.5(V^2 -4) = 4.5 V^2 - 18

3/4.5 = V^2
V = 0.816 m/s

Answer 2

work added up the ramp: force*distance=50*2=100joules
work extraced by gravity= 9(9.8*2*sin19)
final KE=initial KE+workdone-PEgained=1/2 * 9(9.8)2^2+100-9*(9.8)2*sin19

from final KE, solve for final velocity.

Answer 3

push force up ramp = 50 N

work done by push = 50 * 2 = 100 Joules

that 100 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
100 = 57.5 +(1/2)9(V^2 - 2^2)
42.5 = 4.5(V^2 -4) = 4.5 V^2 - 18
60.5 /4.5 =V^2

V = 3.67 m/s
V = 0.816 m/s

Answer 4

push force up ramp = 50 N

work done by push = 50 * 2 = 100 Joules

that 100 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
100 = 57.5 +(1/2)9(V^2 - 2^2)
42.5 = 4.5(V^2 -4) = 4.5 V^2 - 18
60.5 /4.5 =V^2

V = 3.67 m/s