250 W 12 km/h4. Convert yoor answer from 3 to meters per second.
1 answer:
You might be interested in
Answer
given,
mass = 100 kg
acceleration = 10 m/s²
A mass 20 kg slides over 100 kg block
acceleration = 3 m/s²
horizontal friction exerted by the 100 kg block on 20 kg
using newton's second law
F - f = 0
F = f
f = ma
f = 20 × 3
f = 60 N
now net force acting on the 100 kg block
F_net = m a
F_net = 100 x 10
F_net = 1000 N
after 20 kg block falls the acceleration of the bock
F = 1000 +60
F = 1060 N
acceleartion on the block
a = 10.60 m/s²
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