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A 1530 kg rocket is to be launched with an initial upward speed of 54.0 m/s . In

ID: 1879080 • Letter: A

Question

A 1530 kg rocket is to be launched with an initial upward speed of 54.0 m/s . In order to assist its engines, the engineers will start it from rest on a ramp that rises 53 above the horizontal (the figure (Figure 1)). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of 2000 N, and friction with the ramp surface is a constant 500 N.

Part A

How far from the base of the ramp should the rocket start, as measured along the surface of the ramp?

A 1530 kg rocket is to be launched with an initial upward speed of 54.0 m/s . In order to assist its engines, the engineers will start it from rest on a ramp that rises 53 above the horizontal (the figure (Figure 1)). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of 2000 N, and friction with the ramp surface is a constant 500 N.

Part A

How far from the base of the ramp should the rocket start, as measured along the surface of the ramp?

Explanation / Answer

Using law of conservation of energy,

K1 + U1 + W = K2 + U2

0 + mgh + FL = 0.5 mv^2

(mgL sin53) + (Fthrust - ffriction ) L = 0.5 mv^2

L = (1530 x 54^2)/ 2[(1530 x 9.8 x sin53) + (2000 - 500)]

L = 165.55 m

Comment in case any doubt please rate my answer ....

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