A hockey puck with mass 0.160kg is at rest on the horizontal, frictionless surface of a rink. A player applies a force of 0.310N to the puck, parallel to the surface of the ice, and continues to apply this force for 1.60s. What is the (a)position and (b)speed at time end?
To find the position and speed of the hockey puck at the end of 1.60 seconds, we need to use Newton's second law of motion and the equations of motion.
First, let's use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration: F = ma.
In this case, the net force acting on the hockey puck is the force applied by the player, which is 0.310N. The mass of the puck is 0.160kg. Therefore, we can calculate the acceleration of the puck:
a = F / m
a = 0.310N / 0.160kg
a = 1.94 m/s²
Now, we can use the equations of motion to find the position and speed at the end of 1.60 seconds.
(a) Position:
The equation to use is: x = x0 + v0t + (1/2)at²
Since the hockey puck is initially at rest (x0 = 0) and the player applies the force parallel to the ice surface (no vertical movement), we can ignore the vertical components. This means that the initial velocity (v0) is also zero.
x = (1/2)at²
x = (1/2)(1.94 m/s²)(1.60s)²
x ≈ 2.48 meters
Therefore, the position of the hockey puck at the end of 1.60 seconds is approximately 2.48 meters.
(b) Speed:
The equation to use for speed is: v = v0 + at
Since the initial velocity (v0) is zero, the equation simplifies to:
v = at
v = (1.94 m/s²)(1.60s)
v ≈ 3.1 m/s
Therefore, the speed of the hockey puck at the end of 1.60 seconds is approximately 3.1 m/s.