A ball is rolled at a velocity of 25 m/sec. after 13.5 seconds, it comes to a stop . What is the acceleration of the ball?
1 answer:
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Answer:
The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.
Explanation:
Given:
Radius of the curve (R) = 150 m
Maximum speed of truck (v) = 35.5 m/s
Let the coefficient of friction between the roadway and the wheels of the truck be "μ".
As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.
The centripetal force acting on the truck is given as:
Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.
Frictional force is given as:
Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,
Therefore, frictional force,
Now, frictional force = centripetal force
Plug in the given values and solve for 'μ'. This gives,
Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
Learn more about forces:
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