A21.0-kg block is initially at rest on a horizontal surface. A horizontal force of73.0N is required to set the block in motion,
1 answer:
Answer:
a) μs=0.34
b) μk=0.27
Explanation:
Given that
Mass , m = 21 kg
Force required to move the block ,F= 73 N
The force required to keep the block in motion ,F ' = 58 N
Coefficient of the static friction :
The maximum value of the static friction force
fr= μs m g
At rest position
F= fr
73 = μs x 21 x 10 (take g= 10 m/s²)
μs=0.34
Therefore the Coefficient of the static friction will be 0.34.
Coefficient of the kinetic friction ;
Value of the kinetic friction force ,fr= μk m g
F' = Fr
To move the block at constant speed
58 = μs x 21 x 10 (take g= 10 m/s²)
μk=0.27
Therefore the Coefficient of the static friction will be 0.27.
You might be interested in
Answer:
Explanation:
given,
length of ladder = 10 ft
let x be the distance of the bottom and y be the distance of the top of ladder.
x² + y² = 100
differentiating with respect to time we get
..............(1)
when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s
from equation (1)
now,
let the angle between the ladders be θ
y = xtan θ
Answer:
<h2>6 s</h2>Explanation:
The time taken by the cyclist can be found by using the formula
v is the velocity
d is the distance
From the question we have
We have the final answer as
<h3>6 s</h3>Hope this helps you