Question Video: Finding the Motor’s Force of a Car Moving with a Uniform Velocity given the Resistance Force | Nagwa Question Video: Finding the Motor’s Force of a Car Moving with a Uniform Velocity given the Resistance Force | Nagwa

Question Video: Finding the Motor’s Force of a Car Moving with a Uniform Velocity given the Resistance Force Mathematics • Third Year of Secondary School

A car of mass 1.8 metric tons is moving at a constant speed on a horizontal road. If the resistance to the motion is 57.6 kg-wt per tonne of the car’s mass, find the force of its motor.

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Video Transcript

A car of mass 1.8 metric tonnes is moving at a constant speed on a horizontal road. If the resistance to the motion is 57.6 kilogramme-weight per tonne of the car’s mass, find the force of its motor.

To answer this question, let’s recall Newton’s first law of motion. Newton’s first law states that an object at rest remains at rest and an object in motion remains in motion with the same velocity unless acted upon by an external force. This is sometimes alternatively stated that if the net force — that’s the vector sum of all forces acting on the object — is zero, then the velocity of the object is constant.

Now our car is indeed moving at a constant speed. So this means we know that the net force or the vector sum of all forces acting on the car must be equal to zero. So let’s draw a little diagram and see what’s going on.

In a horizontal direction, the car has two forces acting on it. It has a driving force, which acts in the forward direction, and a resistance force. Now I’ve labelled this 𝑅. But we’re told the resistance of the motion is 57.6 kilogramme-weight per tonne. We’re also told the car has a mass of 1.8 metric tonnes.

And so if the resistance to motion is 57.6 kilogramme-weight per tonne of the car’s mass, then for every 1.8 metric tonnes the resistance is 57.6 times 1.8. 𝑅 is therefore equal to 103.68 kilogramme-weights.

Since the speed remains constant, we know that the net force is equal to zero. And since it’s moving to the right in our diagram, 𝐹 minus 𝑅 must be equal to zero. That is, 𝐹 minus 103.68 equals zero. We solve this equation for 𝐹 by adding 103.68 to both sides. So 𝐹 must be equal to 103.68 kilogramme-weight.

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