Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
1 answer:
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Answer:
c. 0.816
Explanation:
Let the mass of car be 'm' and coefficient of static friction be 'μ'.
Given:
Speed of the car (v) = 40.0 m/s
Radius of the curve (R) = 200 m
As the car is making a circular turn, the force acting on it is centripetal force which is given as:
Centripetal force is,
The frictional force is given as:
Friction = Normal force × Coefficient of static friction
As there is no vertical motion, therefore, . So,
Now, the centripetal force is provided by the frictional force. Therefore,
Frictional force = Centripetal force
Plug in the given values and solve for 'μ'. This gives,
Therefore, option (c) is correct.
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
