A hockey player makes a slap shot exerting a constant force of 40.0 newtons on the puck for .2 seconds. What is the magnitude of
1 answer:
Answer:
Magnitude of impulsive force is <u>8 N.s</u>.
Explanation:
Given:
Force acting on the puck is,
Time interval for which the force acts is,
We are asked to find the impulsive force.
Impulsive force acting on a body is the sudden change in the momentum of the body due to the application of a large constant force for a very small interval of time.
Here, the hockey player applies a large force of 40 N for a very short interval of time. So, the impact on the puck is measured as Impulse and is represented by 'J'.
Impulse in terms of applied force and time interval is given as:
Plug in the given values and solve for 'J'. This gives,
Therefore, the magnitude of impulsive force is 8 N.s.
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Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
