A 6.00-kg box is sliding to the right across the horizontal floor of an elevator. The coefficient of kinetic friction between th
e box and the floor is 0.360. Determine the kinetic frictional force that acts on the box when the elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2.
1 answer:
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Angular velocity of the rotating tires can be calculated using the formula,
v=ω×r
Here, v is the velocity of the tires = 32 m/s
r is the radius of the tires= 0.42 m
ω is the angular velocity
Substituting the values we get,
32=ω×0.42
ω= 32/0.42 = 76.19 rad/s
= 76.19× revolution per min
=728 rpm
Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.
Answer: 2940 J
Explanation: solution attached:
PE= mgh
Substitute the values:
PE= 10kg x 9.8 m/s² x 30 m
= 2940 J
Answer:
Acceleration will increase.
Explanation:
The relation between force, mass and acceleration according to the Newton's second law of motion is given as:
F = ma
We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:
Since, F is constant, the acceleration of the truck is inversely proportional to the mass.
There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.
Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.
So, the acceleration of the truck will increase.