4) A 250 gr mass is projected up along a 20° incline. It is observed that as the

Question

4) A 250 gr mass is projected up along a 20° incline. It is observed that as the mass passes a point located 2 m up along the incline, it is moving at a speed of 5 m/sec, and as it passes a point located 4 m up along the incline, it is moving at 3 m/sec. a) What is the normal force of the mass? b) What is the friction force acting on the system? c) How long does the mass take to move from 2 m to 4 m? d) What is the acceleration of the mass? e) What is the coefficient of friction between the mass and the incline?

Explanation / Answer

let's find the retardation up the incline

a= (V^2-U^2) / 2s= (25- 9) /(4 ) = 4m/s^2

a) N ormal force = mgcos theta=(250 x 10^-3) (9.8) cos 20 =2.302 N apprx

b) ma = mgsin theta + force of friction

250 (4) x10^-3 = 10^-3 (250)9.8 sin 20 + force of fruction

force of friction = 1N - 0.8379 N = 0.162 N

c) t= ( v-u)/a = (3-5) / 4= 0.5 seconds

d) a= 4m/s^2

e) Fs= u mgcostheta

0.162 = u ( 250 x 10^-3) (9.8) cos 20

u = 0.162/ 2.3 = 0.07 apprx

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