Find: The magnitude of velocity required at the top of the incline for the monke

Question

Find:

The magnitude of velocity required at the top of the incline for the monkey and sled to reach the Landing Point?

The magnitude of velocity required at the point of release from the spring for the monkey and sled to climb the incline and have sufficient speed to fly off the incline and reach the Landing Point is closest to?

The distance the spring must be compressed for the monkey and sled to have the required speed to fly off the incline and reach the Landing Point is closest to?

At the Landing Point the sled absorbs the vertical component of their velocity while preserving the horizontal component of their velocity. The coefficient of kinetic friction between the sled and the Horizontal Plain required to bring the monkey and sled to a stop at the bluff without plunging into the Bottomless Chasm is closest to?

Monkey on Sled (Friction) Buff Large-X1 Spring X2 -X1 (No Friction) 4 Landing Horizontal Plain Point (Friction) Bottomless Chasm

Explanation / Answer

we assume the verticalvelocity of the sled and the monkey = o when it reached the top of the incline as they do not need any verticalvelocity.

hright to descend h = 21 m

time of descent t = sqrt(2h/2g) = sqrt(2*21/9.8) = 2.07 s

in this time the sled shall travle a distance of x3= 23 m horizontally

horizontal velocity vx = 23/2.07 = 11.59 m/s

horizontal plain x4 = 45 m

final horizontal velocity =0 for the sled to come to stop at the end of x4

acceleration a = -(11.59)^2/2*45 = -1.49m/s/s, acceleration is ove as it is opposite to the motion of the body.

frictional force = ma = 120*1.49

Normal reaction = 120*9.8

co-efficient of friction = 1.49/9.8 = 0.15

angle of the incline arctan(21/60) = 19.3 deg

Normal reaction on the incline = 120*9.8*Cos(19.3) = 1110 N

frictional force on the incline = 0.45*1110 = 500N ( 0.45 co-efficient of friction)

length of the incline = sqrt(60^2+21^2) = 63.57 m

for the sled and the monky to complete the challenge it requires enough energy

1) potential energy to reach height y=21 m

2) KE to have a horizonalt velocity of 11.59 m/s to complete the landing

3) work done against the frictional force up the incline.

1, PE = mgh = 120*9.8*21 = 24696 N

2. KE required = mv^2/2 = 120*(11.59)^2/2 = 8060 N

3. Work done against friction = 500*63.57 = 31784 N

Total energy required to complete the challenge =31784+8060+24696 = 64540 N

This shall come from the compression of the spring = kx^2/2 = 64540

k = 1520 N/m

x = sqrt(64540*2/1520) = 9.22 m

required compression of the spring

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