Question

Find the change in velocity of a 369 g hockey puck subject to the force shown below.

Answer 1

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

$185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]$

$a=501.35[m/s^2]$

Now using the following kinematic equation:

$V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m$

Now replacing the values:

$V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]$