A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of \( 60^\circ \) by a force of 10 N parallel to the inclined surface. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is:
Decompose the weight into components parallel and perpendicular to the inclined plane:
• Component of gravitational force parallel to the plane:
Fgravity,‖ = mg sin θ.
Given m = 1 kg, g = 10 m/s², and sin 60° = √3/2:
Fgravity,‖ = 1 × 10 × (√3/2) = 5√3 N ≈ 8.660 N.
• Component of weight perpendicular to the plane (this is the magnitude of the normal force if no other vertical forces are present):
N = mg cos θ.
Using cos 60° = 1/2:
N = 1 × 10 × (1/2) = 5 N.
Step 2 — Compute the frictional force
The frictional force (kinetic or limiting static, as appropriate) is calculated as:
Ffric = μN.
Substituting μ = 0.1 and N = 5 N:
Ffric = 0.1 × 5 = 0.5 N.
Note: The frictional force remains constant in this scenario because both the coefficient of friction (μ) and the normal force (N) are constant along the incline.
Step 3 — Work done against friction over 10 m
The work done against a constant force is the product of the force and the distance over which it acts (considering the component of force opposite to the direction of motion):
Wfric = Ffric × distance.
Therefore:
Wfric = 0.5 N × 10 m = 5 J.
Step 4 — (Optional) Energy perspective & total work required
If you need to calculate the work required to raise the block against gravity (useful for determining the applied force needed for constant speed motion):
The vertical height gained along the incline is: h = distance × sin θ = 10 × (√3/2) = 5√3 m ≈ 8.660 m.
The increase in potential energy (which represents the work done against gravity) is:
ΔPE = m g h = 1 × 10 × 5√3 = 50√3 J ≈ 86.602 J.
Consequently, if the block is moved upwards at a constant velocity (implying no change in kinetic energy), the total work performed by the applied force is:
Wtotal = ΔPE + Wfric ≈ 86.602 + 5 = 91.602 J.
(However, the specific question requested only the work done against friction, which is 5 J.)
Step 5 — Units, sign convention & interpretation
• Work done against friction is reported as a positive value, signifying the energy expended to overcome frictional resistance — in this case, 5 J.
• If the question asked for the work done by friction itself, the value would be negative (−5 J) because the frictional force opposes the direction of displacement.
• When calculating the work done by an applied force moving an object at constant speed, both the change in potential energy (ΔPE) and the work done against friction (Wfric) must be included, as demonstrated above.
Common mistakes to avoid
• Discrepancies in the value of 'g': the calculation used g = 10 m/s² as specified; using g = 9.8 m/s² would yield slightly different results.
• Incorrect assumption that the normal force equals mg; it is actually mg cos θ.
• Misunderstanding the friction formula; friction is calculated as μN, where μ is the coefficient of friction multiplied by the normal force, not mg directly.
• Omitting the sin θ factor when calculating vertical height from distance along the incline.
Final answer
Work done against friction = 5 J.
Was this answer helpful?
3
Top Questions on General Principles and Processes of Isolation of Elements