Question Video: Calculating the Resistance of Sand to a Body Released from a Given Height That Penetrated the Sand to a Given Depth | Nagwa Question Video: Calculating the Resistance of Sand to a Body Released from a Given Height That Penetrated the Sand to a Given Depth | Nagwa

Question Video: Calculating the Resistance of Sand to a Body Released from a Given Height That Penetrated the Sand to a Given Depth Mathematics • Third Year of Secondary School

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A body of mass 125 kg fell vertically from a height of 112 cm onto a section of sand. It sank 5 cm into the sand before it came to rest. Using the work–energy principle, calculate the resistance of the sand to the body’s motion. Take 𝑔 = 9.8 m/s².

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Video Transcript

A body of mass 125 kilograms fell vertically from a height of 112 centimeters onto a section of sand. It sank five centimeters into the sand before it came to rest. Using the work–energy principle, calculate the resistance of the sand to the body’s motion. Take 𝑔 equal to 9.8 meters per second squared.

We begin by recalling that the work–energy principle states that the change in kinetic energy of an object is equal to the net work done on the object, where work done is equal to force multiplied by distance and kinetic energy is equal to a half 𝑚𝑣 squared, where 𝑚 is the mass and 𝑣 the velocity of the object. We will begin by sketching a diagram to model the situation in this question.

We are told that a body of mass 125 kilograms falls vertically downwards. It falls 112 centimeters, or 1.12 meters, onto a section of sand. It then sank five centimeters, or 0.05 meters, into the sand before it came to rest. We will begin by calculating the velocity of the body just before it hits the sand. This will allow us to calculate the kinetic energy of the body at this point. We will then be able to use this value together with the work–energy principle to calculate the resistance of the sand to the body’s motion. To calculate the velocity 𝑣 in meters per second, we will use the equations of motion or SUVAT equations.

Our value of 𝑠, the displacement, is 1.12 meters. We know that the initial velocity was zero meters per second, and the acceleration due to gravity, 𝑔, is equal to 9.8 meters per second squared. We can use the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have 𝑣 squared is equal to zero squared plus two multiplied by 9.8 multiplied by 1.12. The right-hand side simplifies to 21.952. We might then consider square rooting both sides. However, it is worth noting that our kinetic energy formula contains the term 𝑣 squared. The kinetic energy of the body is therefore equal to one-half multiplied by 125 multiplied by 21.952. Typing this into the calculator gives us 1372. The kinetic energy of the body, as it is about to hit the sand, is 1372 joules.

We can now use the work–energy principle to calculate the force of the body on the sand. The work is 1372 joules and this is equal to 𝐹 multiplied by 0.05. Dividing through by 0.05, we have 𝐹 is equal to 27440. There is a downward force of 27440 newtons on the sand.

We will now clear some space and focus in on the body after it hits the sand to calculate the resistance of the sand to the body’s motion. We have just established that there is a downward force of 27440 newtons based on the motion of the body. There will also be a second downward force equal to the weight of the body. This can be calculated by multiplying 125 by 9.8. It is the mass multiplied by gravity. This is equal to 1225 newtons. The resistance of the sand to the body’s motion will therefore be equal to the sum of these two forces. We need to add 27440 and 1225. This is equal to 28665. The resistance of the sand to the body’s motion is 28665 newtons.

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