Question

a 2.0 kg mass moves along a frictionless horizontal surface at a speed of 5.0 m/s. The mass encounters a 30 degree inclined surface with a constant friction force of 1.5 N. At 1 m high (vertical) the surface levels off and is again frictionless. the mass then encounters a spring with k=10 N/m

a) how far is the spring compressed after the mass comes to rest?

b) how far down the inclined plane will the mass move after bouncing back from the spring if the surface now has a 15 N constant frictional force?

Answer 1

here,

mass ,m = 2 kg

initial horizontal speed , u = 5 m/s

theta = 30 degree

constant friction force , ff = 1.5 N

height , h = 1 m

spring constant , K = 10 N/m

a)

let the spring is compressed x m

using Work energy theorm

- ff * (h /sin(theta)) = 0.5 * K * x^2 + m * g * h - 0.5 * m * u^2

- 1.5 * (1 /sin(30)) = 0.5 * 10 * x^2 + 2 * 9.81 * 1 - 0.5 * 2 * 5^2

solving for x

x = 0.69 m

the compression in the spring is 0.69 m

b)

when constant frictional force , ff2 = 15 N

using Work energy theorm

- ff2 * d = (0 - 0.5 * K * x^2 - m * g * d * sin(theta))

15 * d = (0 + 0.5 * 10 * 0.69^2 + 2 * 9.81 * d * sin(30))

solving for d

d = 0.45 m

the distance traveled is 0.45 m