A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric
1 answer:
Answer:
C. 180 N
Explanation:
= mass of the blocking sled = 60 kg
= Coefficient of kinetic friction = 0.30
= kinetic frictional force acting on the sled
kinetic frictional force acting on the sled is given as
=Force applied
For the sled to move at constant speed, the applied force must be same as the frictional force
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We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
Answer:
422.36 N
Explanation:
given,
time of rotation = 4.30 s
T = 4.30 s
Assuming the diameter of the ring equal to 16 m
radius, R = 8 m
v = 11.69 m/s
now, Force does the ring push on her at the top
N = 422.36 N
The force exerted by the ring to push her is equal to 422.36 N.