A 4.5 kg box slides down a 4.3 m-high frictionless hill, starting from rest, acr

Question

A 4.5 kg box slides down a 4.3 m-high frictionless hill, starting from rest, across a 1.9 m-wide horizontal surface, then hits a horizontal spring with spring constant 500 N/m. The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 1.9 m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.22.

A- What is the speed of the box just before reaching the rough surface?

B- What is the speed of the box just before hitting the spring?

C- How far is the spring compressed?

D- Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

Explanation / Answer

A . by energy conservation , its potential energy will convert into kinetic energy when it comes down

mgh = mv2 /2

v = 2gh = (2 X 9.8 X 4.3) = 9.18 m/s ......Ans

B. just before hitting spring it travels 1.9 m long friction surface ,

f = .mg

a =f / m = .mg / m = .g = 0.22 x 9.8 = 2.156 m/s2

so its speed is v just befire hitting spring then

v2 - u2 = 2as

v2 - 9.182 = 2 (-2.156)x1.9

v2 = 76.08

v = 8.72 m/s     ..............ANS

C . mv2 /2 = kx2 /2

4.5 x 8.722   = 500 X x2

x = 0.8273 m = 82.73 cm

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