Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is
Answer:
10s
Explanation:
Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.
If acceleration is constant, then the velocity is changing by a constant amount.
With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:
where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.
So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.
<u>Verification:</u>
In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.
Recognizing that the acceleration is , the velocity increases by 100 units [m/s] every second.
At time 0[s], the velocity is 0[m/s]
At time 1[s], the velocity is 100[m/s]
At time 2[s], the velocity is 200[m/s]
At time 3[s], the velocity is 300[m/s]
At time 4[s], the velocity is 400[m/s]
At time 5[s], the velocity is 500[m/s]
At time 6[s], the velocity is 600[m/s]
At time 7[s], the velocity is 700[m/s]
At time 8[s], the velocity is 800[m/s]
At time 9[s], the velocity is 900[m/s]
At time 10[s], the velocity is 1000[m/s]
So, indeed, after 10 seconds, the velocity reaches 1000 m/s