Could anyone help me with this question.. plz! A 5.80 kg block is set into motio

Question

Could anyone help me with this question.. plz!

A 5.80 kg block is set into motion up an inclined plane with an initial speed of v1=8.20 m/s. The block comes to rest after traveling d 3.00 m along the plane, which is inclined at an angle of theta=30.0 degree to the horizontal. a. For this motion, determine the change in the block?s kinetic energy. b. For this motion, determine the change in potential energy of the block-Earth system. c. Determine the friction force exerted on the block (assumed to be constant). d. What is the coefficient of kinetic friction?

Explanation / Answer

The total KE at start =0.5 * m * V0^2 = 0.5 * 5.8 * (8.2)^2 = 195 N-m

(a)

As the Final KE=0,

Net change in KE is 195 N-m

(b)

PE at start is zero.

PE at destination = m*g*h= 5.8 * 9.8 * 3 * Sin30= 85.26 N-m

So this is the total Change in PE.

(c)

Ideally, whole of KE should have converted to PE.

So energy lost to friction = KE-PE = 195 - 85.26 = 109.74 N-m.

As energy = F.d,   F*3 = 109.74

So Force of friction F= 109.74 / 3= 36.58N

(d)

For 5.8 Kg at 30 deg, Normal Force N=5.8 *Cos30 * 9.8 = 49.22

So Coef. of Kinetic Friction = 36.58 / 49.22= 0.743

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