A 0.400-kg block is pushed against a 450 N/m spring, compressing it 22.0 cm. Whe

Question

A 0.400-kg block is pushed against a 450 N/m spring, compressing it 22.0 cm. When the block is released, it moves along a frictionless horizontal surface and then up an incline. The angle of the incline is 30.0 degrees and the coefficient of friction with the incline is 0.25. Use the conservation of energy law to find a) the speed of the block just before going up the incline; b) the distance the block travels up the incline; c) the speed of the block when it returns on the horizontal surface; d) the maximum compression of the spring when the block returns to it.

Explanation / Answer

a)

(1/2) mv^2 = (1/2)kx^2

v = sqrt(kx^2/m)

v = sqrt(450*(22/100)^2/0.400)

v = 7.379 m/s

b)

(1/2)mv^2 = mg(d*sin(30)) + um(g*cos(30))d

(1/2)v^2 = g(d*sin(30)) + u(g*cos(30))d

(1/2)7.379^2 = (sin(30) + 0.25*cos(30))gd

10.89 = (sin(30) + 0.25*cos(30))*9.8d

d = 10.89/((sin(30) + 0.25*cos(30))*9.8)

d = 1.5509 m

c)

(1/2)mv^2 = mg(d*sin(30)) - um(g*cos(30))d

(1/2)v^2 = (sin(30) - 0.25*cos(30))gd

(1/2)v^2 = (sin(30) - 0.25*cos(30))*9.8*1.5509

(1/2)v^2 = 4.308769

v = sqrt(4.308769*2)

v = 2.9355 m/s

d)

(1/2) mv^2 = (1/2)kx^2

x = sqrt(mv^2/k)

x = sqrt(0.400*2.9355^2/450)

x = 0.0875 m

x = 8.75 cm

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