Question Video: Power of Motion on Inclined Planes with Resistive Forces | Nagwa Question Video: Power of Motion on Inclined Planes with Resistive Forces | Nagwa

# Question Video: Power of Motion on Inclined Planes with Resistive Forces Mathematics • Third Year of Secondary School

## Join Nagwa Classes

A lorry of mass 3 metric tons with an engine of 52 hp was moving along a section of horizontal road at its maximum speed of 108 km/h. It was then loaded with a weight of 315 kg-wt. After which, it started moving up a road that was inclined to the horizontal at an angle whose sine is 1/17. Given that the resistance force to the lorry’s motion on the inclined road is three times that on the horizontal one, determine the lorry’s maximum speed on the inclined road.

08:25

### Video Transcript

A lorry of mass three metric tons with an engine of 52 horsepower was moving along a section of horizontal road at its maximum speed of 108 kilometers per hour. It was then loaded with a weight of 315 kilogram-weight, after which it started moving up a road that was inclined to the horizontal at an angle whose sine is one seventeenth. Given that the resistance force to the lorry’s motion on the inclined road is three times that on the horizontal one, determine the lorry’s maximum speed on the inclined road.

In order to answer this question, we will need to use the formula that power is equal to force multiplied by velocity. Our standard units for power are watts, for forces, they are newtons, and for velocity, meters per second. In this question, we notice that some of our units are different. So we will need to recall some conversions.

When dealing with power, one horsepower is equal to 735 watts. We know that one ton is equal to 1000 kilograms. We also know that one kilogram-weight is equal to 9.8 newtons. Finally, 3.6 kilometers per hour is equal to one meter per second. We will now use these conversions so that our measurements are in the correct units on the horizontal road and the inclined roads.

The mass of the lorry on the horizontal road is equal to 3000 kilograms as one metric ton is equal to 1000 kilograms. As the lorry is then loaded with a weight of 315 kilogram-weight, its mass on the inclined road is 3315 kilograms. As one horsepower is equal to 735 watts, the power in the engine will be equal to 38220 watts. This is equal to 52 multiplied by 735.

The maximum velocity on the horizontal road was 108 kilometers per hour. Dividing this by 3.6 gives us 30 meters per second. We need to calculate the lorry’s maximum speed on the inclined road. We will call this 𝑉 meters per second. We are also told that the sin of the angle of incline is equal to one seventeenth. If we call this angle 𝛼, sin of 𝛼 is equal to one seventeenth. Finally, we are told that the resistance to the lorry’s motion on the inclined road is three times that on the horizontal one. If we let the resistance on the horizontal road be 𝑅 newtons, then on the inclined road, it is equal to three 𝑅 newtons. We will now clear some space and sketch the two scenarios.

When the lorry is traveling on the horizontal road, we have four horizontal and vertical forces. Vertically downwards, we have the weight of 3000 multiplied by gravity 𝑔. Going vertically upwards, we have the normal reaction force labeled 𝑁. We have a horizontal force 𝐹 acting in the direction of motion and a resistance force 𝑅 acting against the motion. As the lorry is traveling at its maximum velocity, we know the acceleration will be equal to zero. This means that the sum of the net forces vertically and horizontally will equal zero.

Resolving horizontally, this means that 𝐹 minus 𝑅 is equal to zero. Adding 𝑅 to both sides, we see that the force 𝐹 is equal to the resistance force. Using the equation power is equal to force multiplied by velocity, we have 38220 is equal to 𝐹 multiplied by 30. Dividing both sides of the equation by 30 gives us 𝐹 is equal to 1274. The resistance force is, therefore, equal to 1274 newtons. We can now calculate the resistance on the inclined plane by multiplying this number by three. This is equal to 3822 newtons.

We will now sketch the situation on the inclined plane. Once again, the acceleration will be equal to zero as we are interested in the point where the lorry is traveling with maximum velocity. The weight force acting vertically downwards is equal to 3315 multiplied by gravity. We still have a normal reaction force 𝑁 acting perpendicular to the plane. Parallel to the plane, we have the resistance force 𝑅 equal to 3822 newtons acting against the motion and a force 𝐹 acting in the positive direction. Once again, we know that the sum of the net forces acting parallel and perpendicular to the plane are equal to zero.

In order to solve in these directions, we need to find the components of the weight perpendicular and parallel to the plane. We can do this using our knowledge of right angle trigonometry to calculate the value of 𝑥 and 𝑦. In this question, we are only interested in the value of 𝑥 as we will only be resolving parallel to the plane. Using our trig ratios, we know that sin of angle 𝛼 is equal to the opposite over the hypotenuse. This means that sin of 𝛼 in this question is equal to 𝑥 over 3315𝑔. We also know that sin of 𝛼 is equal to one seventeenth. Multiplying both sides of this equation by 3315𝑔 gives us 𝑥 is equal to 1911. The component of the weight force that is parallel to the plane is 1911 newtons. We know that this acts against the direction of motion.

As the sum of the net forces equals zero, 𝐹 minus 𝑅 minus 1911 must equal zero. We know that 𝑅 is equal to 3822. We can substitute this into the equation and then collect like terms. Our equation simplifies to 𝐹 minus 5733 is equal to zero. Adding 5733 to both sides, we see that the positive force is equal to 5733 newtons.

We can now use the fact that power is equal to force multiplied by velocity once again. 38220 is equal to 5733 multiplied by 𝑉. Dividing both sides by 5733 gives us 20 over three or twenty-thirds. This is the maximum velocity of the lorry in meters per seconds. As the first velocity was given in kilometers per hour, it makes sense to convert this answer to kilometers per hour. To do this, we multiply twenty-thirds by 3.6. The lorry’s maximum velocity or speed on the inclined road is 24 kilometers per hour.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions