Question

A block with a weight of 3.7 N is launched along a slope upwards by a spring with k = 2.25 kN / m. The slope makes an angle of 30° with the horizontal and has a vertical height of 0.75 m (measured from the starting point of the block). When the block is released, the spring is pressed 0.10 m. The kinetic coefficient of friction between the block and the slope is 0.30. What is the size of the block speed at the top of the slope?

Answer 1

M*g = 3.7 N. = Wt. of the block.
M = 3.7/9.8 = 0.378 kg.

Fn = 3.7*Cos30 = 3.2 N. = Normal force.

Fk = u*Fn = 0.3 * 3.2 = 0.96 N. = Force of kinetic friction.

d = h/sin30 = 0.75/sin30 = 1.5m. = Distance to top of incline.

KE = PE-Fk*d.
0.5M*V^2 = Mg*h-Fk*d
0.5*0.378V^2 = 3.7*0.75-0.96*1.5.
V = ?.

Answer 2

I used the formula of conservation of mechanical energy. Where my spring is PE at start and that's equal to PE at the top + KE + Friction F