A 15.0-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 15.0-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 m/s in 0.600 s. In the process, the spring is stretched by 0.210 m. The block is then pulled at a constant speed of 5.00 m/s, during which time the spring is stretched by only 0.0500 m. (a) Find the spring constant of the spring. N/m (b) Find the coefficient of kinetic friction between the block and the table.

Explanation / Answer

m = mass = 15 kg

speed = 5m/s

t = time = 0.600 s

block accceleration, a = 5/ 0.600 = 8.33 m/s2

accelerating force , F = m*a

= 15* 8.33 = 124.95 N

friction force will also stretch the spring but in opposite direction. this is the force which stretched by constant speed. so from Hook's law

f ( friction force) = -kx ( k = spring constant , here x = 0.0500m)

[ here -ve sign for direction]

f = k* 0.0500

total stretching force = F+ f

k*x' = 124.95 + 0.0500 *k [ here x' = 0.210m]

0.210 k - 0.0500 k = 124.95

k = 780.83 N/m

ii) friction force f = uk*R

here uk = cofficient of kinetic friction, R(normal reaction of block) = mg , f = k*0.0500

so k*0.0500 = uk*m*g

uk = 780.93 *0.0500 / 15*9.8

uk = 0.265

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