A 10-kg block is suspended in air by a single string that passes through a pulley and is attached on the other side to a 35-kg b
1 answer:
Answer:
0.286
Explanation:
Let g = 10m/s2, and assume the pulley is frictionless, the weight by the 10kg block would exert a force on the 35kg block on the ground
Weight of the 10 kg-block = mg = 10 * 10 = 100 N
This force is balanced by static friction force from the ground due to the normal force of the 35kg block. So the friction force would also be 100N
The weight and normal force of the 35kg block is N = Mg = 35*10 = 350 N
The coefficient of the friction force if friction force divided by normal force:
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Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed
Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s