Question

A hockey puck with mass 0.160 kg is at rest at the origin (x=0) on the horizontal, frictionless surface of the rink. At time t=0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; she continues to apply this force until t=2.00 s. (a) What are the position and speed of the puck at t=2.00 s? (b) If the same force is again applied at t=5.00 s, what are the position and speed of the puck at t=7.00 s?​

Answer 1

Answer:

The answer is a) 3.125 m & 3.125 m/s and b) 21.875 m & 6.25 m/s.

Explanation:

a) Given information:

Initial velocity (u) = 0 m/s

Mass (m) = 0.16 kg

Force (F) = 0.25 N

t = 2 s

Since,

Force (F) = Mass (m) x acceleration (a)

Hence,

0.25 = 0.16 x a

a = 0.25/0.16 = 1.5625 $m/s^{2}$

According to first law of motion, position (s) is given by:

$s = ut + \frac{1}{2} at^{2}$

Hence, distance travelled till t= 2s:

s = 0 + (0.5 x 1.5625 x 4) = 3.125 m

Velocity at t = 2 s:

v = u + a.t

v = 0 + 1.5625 x 2 = 3.125 m/s

b) Since there is no acceleration between 2s and 5s, the velocity will remain same.

Hence, distance travelled from 2s to 5s = v.t = 3.125 x 3 = 9.375 m

From t = 5 s to t = 7 s:

Initial velocity (u) = 3.125 m/s

a = 1.5625 $m/s^{2}$

Distance travelled:

$s = ut + \frac{1}{2} at^{2}$

s = (3.125 x 2) + (0.5 x 1.5625 x 4) = 9.375 m

Hence,

total displacement = 3.125 + 9.375 + 9.375 = 21.875 m

Final velocity v:

v = u + a.t

v = 3.125 + (1.5625 x 2) = 6.25 m/s

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