### 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.