David, who is 65 kg, is driving his car at a speed of 25 m/s. He sees a dog crossing the road and so he steps on the brake to avoid hitting it. His seatbelt brings his body to a stop in 0.4 s. What magnitude of force does the seatbelt exert on him?
Well, let me put on my physics clown nose for this one!
So, David weighs 65 kg and his car is cruising at 25 m/s. Suddenly, a dog appears and David hits the brakes. But we're not discussing the dog's speed here, we're talking about the force exerted on David by his seatbelt.
To figure that out, we can use the famous equation:
force = mass × acceleration.
However, in this case, we can use a little trick. David's body comes to a stop due to the seatbelt in 0.4 seconds. So we can actually find the acceleration by dividing the change in velocity by the time it takes:
acceleration = (final velocity - initial velocity) / time.
David's initial velocity is 25 m/s, and his final velocity is 0 m/s (since he comes to a stop), and the time is 0.4 seconds. Plugging those values in, we get:
acceleration = (0 - 25) m/s / 0.4 s.
That gives us a dashing acceleration of -62.5 m/s².
Now, to find the force, we can plug this acceleration and David's mass (65 kg) into the equation:
force = mass × acceleration.
So, force = 65 kg × (-62.5 m/s²), which gives us... a staggering -4062.5 Newtons!
But keep in mind, this negative sign simply indicates that the force acts in the opposite direction of David's motion. So, the seatbelt exerts a force of 4062.5 Newtons on David, trying to prevent his forward motion.
To find the magnitude of force exerted by the seatbelt on David, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
In this case, since David's body comes to a stop, the acceleration experienced by him can be calculated using the equation:
a = Δv / Δt
Where Δv is the change in velocity and Δt is the change in time.
First, let's find the change in velocity:
Δv = final velocity - initial velocity
Since David's body comes to a stop, the final velocity is 0 m/s. The initial velocity is 25 m/s. Therefore:
Δv = 0 - 25 = -25 m/s
Next, let's calculate the change in time:
Δt = 0.4 s
Now, we can find the acceleration:
a = Δv / Δt
a = -25 m/s / 0.4 s
a = -62.5 m/s²
Since the acceleration is negative, it indicates a deceleration.
Finally, we can calculate the magnitude of the force exerted by the seatbelt on David by substituting the values into Newton's second law:
F = m * a
F = 65 kg * -62.5 m/s²
F = -4062.5 N
Therefore, the magnitude of force exerted by the seatbelt on David is 4062.5 Newtons (N).
To find the magnitude of force exerted by the seatbelt on David, we can use the equation:
Force (F) = mass (m) * acceleration (a)
First, let's calculate the acceleration of David's body when it comes to a stop. We can use the equation:
acceleration (a) = change in velocity / time
David's initial velocity is 25 m/s, and since he comes to a stop, the change in velocity is 25 m/s. The time taken to stop is given as 0.4 s.
a = (0 - 25 m/s) / 0.4 s = -62.5 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
Now that we have the acceleration, we can calculate the force exerted by the seatbelt:
F = m * a
David's mass is given as 65 kg, so:
F = 65 kg * (-62.5 m/s²) = -4062.5 N
The negative sign indicates that the force is in the opposite direction to the initial motion. Thus, the magnitude of the force exerted by the seatbelt on David is 4062.5 Newtons.
The velocity changes by -25 m/s in 0.4 s, so the acceleration is
25/.4 m/s^2 = 62.5 m/s^2
Since F=ma, the force exerted is
65 kg * 62.5 m/s^2 = 4062.5 N