A man with mass m =96.4 kg (weight = 212.1 lbs is walking towards a patio at 2.00 m/s (4.47 mph) when, at the last moment, he realizes the sliding glass door is closed. He puts out his hand suddenly to stop himself, and comes to a halt in 90.0ms. Find the magnitude of the average force exerted on the door by his hand.
Use the relation
Impulse = momentum change,
which means
Force * Time = Mass * Velocity
or just
F = M*V/T
How do I find the time? How is 90.0 m/s converted to seconds? Isn't that what I need to plug into the quation?
The number you were given for time T is 90 milliseconds, which is 0.09 s.
The units of time are not m/s (meters per second)
Oh gosh, can't believe I misread that. Thank you.
To find the magnitude of the average force exerted on the door by the man's hand, we need to use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration. In this case, the man comes to a halt, so his final velocity is 0 m/s. We can calculate the acceleration using the equation:
a = (vf - vi) / t
Where:
- a is the acceleration
- vf is the final velocity (0 m/s in this case)
- vi is the initial velocity (2.00 m/s in this case)
- t is the time taken to stop (90.0 ms = 0.09 s)
Substituting the values, we have:
a = (0 - 2.00) / 0.09
= -2.00 / 0.09
= -22.22 m/s^2
Since the man is exerting the force on the door, the force exerted on the door is equal in magnitude but opposite in direction to the force exerted on the man. Therefore, the force exerted on the man is:
F = m * a
Substituting the mass of the man (96.4 kg) and the acceleration (-22.22 m/s^2), we have:
F = 96.4 * (-22.22)
= -2144.6488 N
Taking the magnitude of the force, we have:
Magnitude of force = |F| = |-2144.6488| = 2144.6488 N
Therefore, the magnitude of the average force exerted on the door by the man's hand is approximately 2144.65 N.