Physics: Finding initial speed given weight, distance and coefficient of friction.?

Question

A 14 kg block is at rest on a level floor. A 400 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty?

_____m/s

Answer 1

The velocity of the block and the putty as they slide 15 cm along the floor must first be determined before the initial velocity of the putty can be calculated.

Thus being said, the working formula is

Vf^2 - V^2 = 2as  --- call this Equation 1

where

Vf = final velocity of block and putty = 0 (as they will stop)

V =  velocity of the putty and block before sliding

a = acceleration of the putty and block

s = distance travelled by putty and block before stopping = 15 cm = 0.15 m. (given)

The acceleration of the block and the putty is,

a = - (μk * g)

where

μk = coefficient of friction = 0.4 (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Hence,

a = - 0.4 * 9.8 = - 3.92 m/sec^2  --- the negative sign attached to the acceleration implies that the block and putty were slowing down as they were skidding to a stop.

Substituting appropriate values in Equation 1,

0 - V^2 = 2(-3.92)(0.15)

V^2 = 1.176

V = 1.08 m/sec.

The next working formula is the law of conservation of momentum, i.e.,

MbVb + MpVp = (Mb + Mp)(1.08)

where

Mb = mass of the block = 14 kg (given)

Vb = initial velocity of block = 0 (at rest)

Mp = mass of putty = 400 g = 0.4 kg (given)

Vp = initial velocity of putty

Substituting values,

14(0) + 0.4(Vp) = (14.4)(1.08)

Solving for Vp,

Vp = 38.88 m/sec.