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.
To find the magnitude of the average force exerted on the door by the man's hand, we can use Newton's second law of motion, which states that the force (F) exerted on an object is equal to the mass (m) of the object multiplied by its acceleration (a).
Given:
Mass of the man, m = 96.4 kg
Velocity of the man, v = 2.00 m/s
Time taken to come to a halt, t = 90.0 ms = 90.0 × 10^(-3) s
To calculate the acceleration, we can use the equation:
a = (change in velocity) / (time taken)
First, we need to convert the time from milliseconds to seconds:
t = 90.0 × 10^(-3) s
Now, let's calculate the change in velocity:
change in velocity = final velocity - initial velocity
= 0 m/s - 2.00 m/s
= -2.00 m/s
Now, we can calculate the acceleration:
a = (-2.00 m/s) / (90.0 × 10^(-3) s)
Finally, we can calculate the average force exerted on the door by the man's hand:
F = m * a
Substituting the given values:
F = 96.4 kg * [(-2.00 m/s) / (90.0 × 10^(-3) s)]
Calculating this expression will give you the magnitude of the average force exerted on the door by the man's hand.
To find the magnitude of the average force exerted on the door by the man's hand, we can use Newton's second law of motion, which states that the force (F) is equal to the mass (m) multiplied by the acceleration (a). The equation is written as F = ma.
First, let's convert the man's speed from mph to m/s. We know that 1 mile is equal to 1.60934 kilometers (km), and 1 hour is equal to 3600 seconds (s):
Speed in m/s = (Speed in mph) × (1.60934 km/mile) × (1000 m/km) / (3600 s/hour)
Speed in m/s = 4.47 mph × 1.60934 km/mile × 1000 m/km / 3600 s/hour
Speed in m/s ≈ 2.00 m/s
Now, let's calculate the acceleration (a). We can use the equation a = Δv / t, where Δv is the change in velocity and t is the time taken:
Acceleration (a) = (Final velocity - Initial velocity) / Time
Since the man is coming to a halt, the final velocity is 0 m/s:
Acceleration (a) = (0 m/s - 2.00 m/s) / 90.0 ms
Acceleration (a) = (0 - 2.00) / (90.0 × 10^(-3)) m/s^2
Acceleration (a) = -2.00 / 0.0900 m/s^2
Acceleration (a) ≈ -22.2 m/s^2
Now that we have the mass (m) of the man and the acceleration (a), we can find the magnitude of the average force (F) using Newton's second law:
Magnitude of average force (F) = mass (m) × acceleration (a)
Magnitude of average force (F) = 96.4 kg × -22.2 m/s^2
Magnitude of average force (F) ≈ -2139.28 N
Therefore, the magnitude of the average force exerted on the door by the man's hand is approximately 2139.28 Newtons (N). Note that the negative sign indicates that the force is in the opposite direction to the motion.