Calculating Force Based On Newton’s Second Law Of Motion (2)
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Question 1. The engine of a vehicle moves it forward with a force of 9600N against a resistive force of 2200N. If the mass of the vehicle is 3400kg, calculate the acceleration produced. Solution: Note: the net force with which the vehicle accelerated is the difference between the two forces, i.e. 9600N – 2200N = 7400 Newtons Mass = 3400kg, acceleration = ? Therefore, using the formula for force, F = ma 7400 = 3400 x a a = 7400/3400 = approx. 2.18m/s2 Question 2. A net force of 15N acts upon a body of mass 3kg for 5s, calculate the change in the speed of the body. Solution: Given: Force = 15N, mass = 3kg, time = 5s, ∆v = ? Using the force formula, F= m x ∆V/t Restating the formula, we have ∆v = Ft/m ∆v = 15 x 5/3 = 25m/s Question 3. A bob of a simple pendulum has a mass of 0.02kg. Determine the weight of the bob. [g = 10ms-2] Solution: Given: mass = 0.02kg, acceleration due to the force of gravity = 10ms-2 Using the formula for force, F = mg, where g (acceleration due to the force of gravity replaces a (normal acceleration), Weight, W = mg Therefore, W = 0.02 x 10 = 0.2 Newtons Question 4. What change in velocity would be produced on a body of mass 4 kg if a constant force of 16 N acts on it for 2 s? Solution: Given: Force = 16 Newtons, mass = 4 kg, time = 2 s, change in velocity (∆v) = ? Using the formula for force, F = m x ∆V/t 16 = 4 x ∆V/2 Restating the equation, 16 x 2/4 =∆v ∆v = 8 m/s Question 5. A force of 16 N applied to a 4.0kg block that is at rest on a smooth, horizontal surface. What is the velocity of the block at t = 5 seconds? Solution: Using the formula for force, F = m x ∆V/t or F = m x (v – u)/t Given: m = 4.0 kg, final velocity v = ?, initial velocity u = 0, time = 5 seconds. Therefore, 16 = 4.0 x (v – 0)/5 16 = 4.0v/5 4.0v = 80 v = 80/4 = 20 Velocity of the block is 20 m/s See more calculations of force based on Newton’s second law of motion. |