A plane traveling 40 m/s drops a box from a height of 186 m. Where does the box strike the ground relative to the point directly
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Answer:
The skater covers a distance of <u>15 m</u> before stopping.
Explanation:
Let the distance traveled before stopping be 'd' m.
Given:
Mass of the skater (m) = 90 kg
Initial velocity of the skater (u) = 12.0 m/s
Final velocity of the skater (v) = 0 m/s (Stops finally)
Coefficient of kinetic friction (μ) = 0.490
Acceleration due to gravity (g) = 9.8 m/s²
Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.
Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.
Frictional force is given as:
Where, 'N' is the normal force acting on the skater. As there is no vertical motion,
∴
Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:
Now, change in kinetic energy is given as:
Therefore, from work-energy theorem,
Hence, the skater covers a distance of 15 m before stopping.