a 50 kg cart rests on a 37 degree incline against a compressed spring of force c

Question

a 50 kg cart rests on a 37 degree incline against a compressed spring of force constant 1.0x10^5 N/m. Initially, the spring is compressed 0.20 m. When released, the spring launches the cart up the incline. Determine the cart's speed when it is 2.0 m from its starting position. A 100 N friction force opposes its motion along the incline. Thanks for any help I could get. If someone could check my work.
[(0.5)(1.0x10^5N/m)(0.20^2)]+[Wnet] = [(0.5)(50 kg)(v^2)]
[2000 J] + [(m*g*sin(37)*(2.0 m)*cos(180)]+[(100 N)*(2.0 m)*cos(180)] = [(0.5)*(50 kg)*(v^2)]
[2000 J]-[590 J]-[200 J] = (25 kg)*v^2
1210 J = (25 kg)*v^2
v^2= 48.4
v=6.957m/s = 7.0 m/s
If this is wrong could someone correct my work and show me how to do it right.

Explanation / Answer

Your Answer is Correct as you apply law of Conserbation of Work and Energy

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