Question

A hockey puck with a mass of 0.18 kg is at rest on the horizontal frictionless surface of the rink. A player applies a horizontal force of 0.5 N to the puck. Find the speed and the traveled distance 5 s later.

Answer 1

Answer:

1) The speed after 5 seconds is 13.$\bar 8$ m/s

2) The distance traveled after 5 seconds is 34.7$\bar 2$ meters

Step-by-step explanation:

1) The given parameters are;

The mass of the hockey puck, m = 0.18 kg

The horizontal force applied by the player, m = 0.5 N

The time at which the speed is calculated, t = 5 seconds after the force is applied

Therefore;

From the equation for the applied force, F = m × a. where a represents the acceleration, by substituting the known values get;

0.5 = 0.18 × a

∴ a = 0.5/0.18 = 50/18 = 25/9 = 2.$\bar 7$

a = 2.$\bar 7$ m/s²

The speed after 5 seconds is given from the kinematic equation, v = u + a·t

Where;

v = The final velocity

u = The initial velocity = 0 m/s

t = The time = 5 seconds

By substitution of the known values, we get;

v = 0 + 2.$\bar 7$ × 5 = 13.$\bar 8$

The speed after 5 seconds = 13.$\bar 8$ m/s

2) The distance traveled, s, after 5 seconds is given by the equation, s = u·t + 1/2·a·t²

From which we have;

s = 0 × 5 + 1/2 × 2.$\bar 7$ × 5² = 34.7$\bar 2$

The distance traveled after 5 seconds = 34.7$\bar 2$ meters