A 3.6kg mass is released from rest and begins to slide down an 8.0m long frictio

Question

A 3.6kg mass is released from rest and begins to slide down an 8.0m long frictionless incline. When the block has slid 5.3m, a persongrabs a cord attachedto the mass and begins to pullthe cord parallel to the incline. By pulling on the cord, the person is able to slow the mass such that it comes to a stop just as it reaches the end of the 8.0 long incline. Assuming the tension in the cord is constant, find the tension in the cord.

I calculated T= 40.86N Can someone confirm they got the same answer?

Explanation / Answer

acceleration before applying force = g * sin(theta)

acceleration before applying force = 9.8 * sin(17)

velocity after 5.3 m

v^2 = u^2 + 2as

v^2 = 0 + 2 * 9.8 * sin(17) * 5.3

v = 5.51 m/s

let the tension be T so,

force equation will be

T - mg * sin(theta) = ma

a = (T - mg * sin(theta)) / m

a = (T - 3.6 * 9.8 * sin(17)) / 3.6

v^2 = u^2 = 2as

0 = 5.51^2 + 2 * ((T - 3.6 * 9.8 * sin(17)) / 3.6) * (8 - 5.3)

T = 9.925 N

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