Calculate work done by forces using the work equation W = F·d·cos(θ). Determine work from force, displacement, and angle with step-by-step physics solutions for various mechanical scenarios.
Problem: A person pushes a box with a force of 50 N at an angle of 30° above the horizontal over a displacement of 8 meters. How much work is done?
Solution: Using W = F·d·cosθ
W = 50 × 8 × cos(30°) = 50 × 8 × 0.8660 = 346.4 J
The horizontal component of the force (43.3 N) does the work, while the vertical component (25 N) counteracts some of the box's weight.
Problem: A weightlifter lifts a 100 kg barbell vertically upward a distance of 2.2 meters. How much work is done against gravity?
Solution: The force required equals the weight (mg). F = 100 × 9.81 = 981 N. Since lifting is vertical, θ = 0°.
W = 981 × 2.2 × cos(0°) = 981 × 2.2 × 1 = 2158.2 J
At θ = 0°, cos(0°) = 1, so the formula simplifies to W = F·d. This is approximately 2.16 kJ of work.
Problem: A rope pulls a sled with a force of 200 N at an angle of 40° above the horizontal for 15 meters. Find the work done by the rope.
Solution: Using W = F·d·cosθ
W = 200 × 15 × cos(40°) = 200 × 15 × 0.7660 = 2298.1 J
The horizontal component (153.2 N) pulls the sled forward, while the vertical component (128.6 N) reduces the normal force and friction.
Problem: A crate is pushed with a force of 80 N at an angle of 20° below the horizontal across a rough floor for 12 meters. The kinetic friction force is 30 N. What is the net work done?
Solution: Horizontal component of applied force = 80 × cos(20°) = 75.18 N. Net horizontal force = 75.18 - 30 = 45.18 N.
Net work = 45.18 × 12 = 542.2 J
The work done by the applied force is 80 × 12 × cos(20°) = 902.1 J, but friction does -360 J of negative work, resulting in 542.1 J of net work.
Where W is work (Joules), F is the applied force (Newtons), d is the displacement (meters), and θ is the angle between the force vector and the displacement vector.
Use this form when you know the work done and the displacement, and need to find the applied force.
Use this form when you know the work done and the applied force, and need to find the displacement.
Work is the transfer of energy when a force moves an object through a displacement. Only the component of the force parallel to the displacement does work. The SI unit of work is the Joule (J), equivalent to one Newton-meter (N·m).
The angle between the force and displacement vectors is critical. When θ = 0° (force parallel to motion), cos(0°) = 1 and work is maximized. When θ = 90° (force perpendicular), cos(90°) = 0 and no work is done, even if the force is large — like carrying a box horizontally.
When the force has a component in the direction of motion (0° ≤ θ < 90°), work is positive and energy is transferred to the object. When the force opposes motion (90° < θ ≤ 180°), work is negative and energy is removed — like friction slowing a sliding box.
Any force at an angle can be split into horizontal (F·cosθ) and vertical (F·sinθ) components. Only the component parallel to the displacement contributes to work. This is why pushing at a downward angle on a sled makes it harder — the vertical component increases friction.
⚠️ Important Note: This calculator assumes the force is constant over the displacement. In real-world scenarios, forces often vary with position, angle, or time. The work-energy theorem still applies, but calculating work for variable forces requires integration. For accurate results in dynamic systems, consider using calculus or simulation tools.
The concept of work in physics is more specific than the everyday meaning. In physics, work is done only when a force causes a displacement in the direction of that force. If you hold a heavy box stationary, you may feel like you're doing work, but physically — if there is no displacement — no work is done on the box.
The work equation W = F·d·cos(θ) captures three essential elements:
Understanding work is fundamental to mechanics because it bridges forces and energy. The work-energy theorem (Wnet = ΔKE) is one of the most powerful tools in physics, allowing you to analyze motion without needing to track forces over time — just the net work done determines how the kinetic energy changes.
When a force is applied at an angle, resolving it into horizontal and vertical components is essential for analyzing the physics accurately:
Real-world application: When pushing a lawn mower, you push downward and forward (angle below horizontal). The downward component increases the normal force, which increases friction — making it harder to push. In contrast, pulling a sled upward reduces the normal force, decreasing friction and making it easier to move. Understanding these force components helps optimize efficiency in mechanical systems.
Our calculator automatically computes both force components alongside the work result, giving you a complete picture of the forces involved.