Question

A horizontal force of 25 N is applied to a 4-kg block, which is initially at

rest on a horizontal bench. The coefficient of
kinetic friction between the block and the surface
of the bench is 0.35. The block is pushed a
distance of 3m. (5-points) Find, A) The external
work done on the block-bench system. (1-points)
B) The energy dissipated by friction. (1.5-points)
C) The kinetic energy of the block after it has
been pushed by 3 m. (1.5-points) D) The speed of
the block after it has been pushed 3 m. (1-points)
IN, 25 N |

Answer 1

The speed of the block after it has been pushed 3 m is approximately 6.12 m/s.

Step-by-step explanation:

Let's go step by step to solve each part of the problem:

A) The external work done on the block-bench system:

The work done by a force is given by the formula: Work = Force × Distance × cos(θ), where θ is the angle between the force and the direction of displacement.

In this case, the force applied is horizontal, so the angle between the force and displacement is 0 degrees. Therefore, cos(0°) = 1.

Work = Force × Distance × cos(0°) = 25 N × 3 m × 1 = 75 J (joules).

B) The energy dissipated by friction:

The energy dissipated by friction can be calculated using the formula: Energy Dissipated = Friction Force × Distance.

The friction force can be calculated using the formula: Friction Force = coefficient of kinetic friction × Normal Force.

The normal force is the force exerted by the surface perpendicular to the contact surface, which is equal to the weight of the block (mg).

Friction Force = μ × m × g, where μ is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity.

Friction Force = 0.35 × 4 kg × 9.8 m/s² = 13.72 N.

Energy Dissipated = Friction Force × Distance = 13.72 N × 3 m = 41.16 J.

C) The kinetic energy of the block after it has been pushed by 3 m:

The initial kinetic energy is 0 because the block is initially at rest.

The final kinetic energy can be calculated using the formula: Kinetic Energy = 0.5 × mass × velocity².

Since the block has been pushed, its velocity is not zero anymore. We'll calculate the velocity in part D and then use it to find the kinetic energy.

D) The speed of the block after it has been pushed 3 m:

To find the speed, we can use the work-energy principle. The work done on the block is equal to the change in its kinetic energy.

Work = Change in Kinetic Energy

75 J = 0.5 × 4 kg × (final velocity)² - 0.5 × 4 kg × (initial velocity)²

Since the block is initially at rest (initial velocity = 0), the equation simplifies to:

75 J = 0.5 × 4 kg × (final velocity)²

Solving for the final velocity:

(final velocity)² = (2 * 75 J) / 4 kg = 37.5

final velocity = √37.5 ≈ 6.12 m/s.

So, the speed of the block after it has been pushed 3 m is approximately 6.12 m/s.