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; he continues to What is the position of the puck at t=2.00 s ?

Answer 1

The position of the puck at t=2.00s is 3.125 m.

Step-by-step explanation:

To determine the position of the puck at t=2.00s, we need to apply the equations of motion. Since the puck is at rest initially, its initial velocity (v0) is 0. Applying Newton's second law (F = ma), we can calculate the acceleration (a) by dividing the force (F) applied by the mass (m) of the puck. Using the kinematic equation x = x0 + v0t + (1/2)at2, we can then calculate the position (x) of the puck at t=2.00s.

Given:

• Mass of the puck (m) = 0.160 kg
• Force applied (F) = 0.250 N
• Time (t) = 2.00 s

First, calculate the acceleration (a):

• a = F/m = 0.250 N / 0.160 kg = 1.5625 m/s2

Next, calculate the position (x):

• x = x0 + v0t + (1/2)at2
• Since the initial velocity (v0) is 0 and the puck is at the origin (x0 = 0), the equation simplifies to
• x = (1/2)at2 = (1/2) * 1.5625 m/s2 * (2.00 s)2 = 3.125 m

Therefore, the position of the puck at t=2.00s is 3.125 m.

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