A 12 kg block sliding along a smooth (frictionless) horizontal surface goes over

Question

A 12 kg block sliding along a smooth (frictionless) horizontal surface goes over a bump on that surface. The top of the bump is 1.2 meters higher than the rest of thesurface, and the bump has a radius of curvature of 1.7 meters. When the block is at the top of the bump, it is traveling at 2.5 m/s. After going over the bump, the block hits a patch of friction (Uk = 0.65) and slides to a stop.

(a) Find the magnitude of the normal force acting on the box when it is at the top of the bump.

(b) Find the speed of the block before it reached the bump, while it was still traveling along the flat surface

(c) Find the distance the block slides along the friction patch before coming to a stop

Explanation / Answer

a)

at the top of bump

Fn = normal force in Upward direction

W = weight of block in down direction

vt = speed at the top = 2.5 m/s

r = radius of curvature = 1.7 m

force equation is given as

W - Fn = m vt2/r

mg - Fn = m vt2/r

(12 x 9.8) - Fn = (12) (2.5)2/1.7

Fn = 73.5 N

b)

vb = speed before bump

h = height of the bump = 1.2 m

using conservation of energy

kinetic energy before bump = kinetic energy at top of bump + potential energy

(0.5) m vb2 = (0.5) m vt2 + mgh

vb2 = vt2 + 2gh

vb2 = (2.5)2 + 2 (9.8 x 1.2)

vb = 5.5 m/s

c)

vi = initial speed = 5.5 m/s

vf = final speed = 0 m/s

a = acceleration due to friction = - uk g = - (0.65) (9.8) = - 6.4 m/s2

d = distance travelled = ?

using the equation

vf2 = vi2 + 2 a d

(0)2 = (5.5)2 + 2 (- 6.4) d

d = 2.4 m

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