Question:
A box with a mass of 2 kg is initially at rest on a frictionless surface. A constant force of 10 N is applied horizontally to the box, displacing it a distance of 5 meters. The box then encounters a rough surface with a coefficient of kinetic friction of μ=0.2, which acts over a distance of 3 meters. Calculate the work done by the applied force, the work done by the frictional force, and the total work done on the box.
(Assume g=9.8 m/s² for acceleration due to gravity. Round your answers to two decimal places)
Answer:
To calculate the work done by each force, we will use the formula: Work = Force × Distance × cos(θ)
Given: Force (F) = 10 N Distance (d) = 5 m θ = 0° (the force and displacement are in the same direction)
Using the formula, we find: Work by applied force = F × d × cos(θ) = 10 N × 5 m × cos(0°) = 10 N × 5 m × 1 = 50 J
Therefore, the work done by the applied force is 50 Joules.
To calculate the work done by the frictional force, we need to find the frictional force first:
Given: Coefficient of kinetic friction (μ) = 0.2 Mass of the box (m) = 2 kg Acceleration due to gravity (g) = 9.8 m/s²
Frictional force (F_f) = μ × Normal force
Since the box is on a frictionless surface initially, the normal force is equal to the weight of the box:
Normal force (N) = m × g = 2 kg × 9.8 m/s² = 19.6 N
Frictional force (F_f) = μ × N = 0.2 × 19.6 N = 3.92 N
Now, let's calculate the work done by the frictional force:
Given: Distance (d) = 3 m θ = 180° (the force and displacement are in opposite directions)
Using the formula, we find: Work by frictional force = F_f × d × cos(θ) = 3.92 N × 3 m × cos(180°) = 3.92 N × 3 m × (-1) = -11.76 J
Note: The negative sign indicates that the work done by the frictional force is negative, representing energy loss due to friction.
Thus, the work done by the frictional force is -11.76 Joules.
Total work done = Work by applied force + Work by frictional force = 50 J + (-11.76 J) = 38.24 J
Therefore, the total work done on the box is 38.24 Joules.