9. Work is done when a weightlifter holds a barbell still above his head A True B False 8. You push against the back of your friend's car that is stuck. You push and become very tired. If the car does not move, did you do work? A Yes, work was done B No, work wasn't done 9. A mass is lifted upward with a force of 49N, at a constant speed, to a height of 10 m. Calculate the work done by the lifting force. A 490 Joules B 50 Joules C 2 Joules D 49 Newtons 10. Calculate the work done by gravity when a 10kg mass is pulled across a smooth horizontal floor at a constant speed. A 98 Joules B 10 Joules C 0 Joules D 98 Newtons 11. Which requires more work: lifting a 50.0 kg crate a vertical distance of 2.0 meters or lifting a 25.0 kg crate a vertical distance of 4.0 meters? A both require the same amount of work. B lifting the 50 kg crate C too heavy, we need Superman! D lifting the 25 kg crate 12. A woman lifts her 100 N child up 1 meter and carries her a distance of 10 meters to her bedroom. How much work doos the woman do? A 2000 J BOJ C 100J D 1000J 13. If 2.0J of work is done in raising a 0.18kg apple, how far was it lifted? A 1.13 m B 90 m C 360 m D 1.76 m

## Step1<br />For question 9, work is defined as the product of force and the distance over which it is applied. If the weightlifter is holding the barbell still, there is no displacement, and hence no work is done.<br /><br />## Step2<br />For question 8, work is done when a force causes an object to move. If the car does not move, no work is done.<br /><br />## Step3<br />For question 9, work is calculated as the product of force and distance. The force is 49 N and the distance is 10 m. Therefore, the work done is \(49 \times 10 = 490\) Joules.<br /><br />## Step4<br />For question 10, if the mass is moved horizontally at a constant speed, the force of gravity acts perpendicular to the direction of motion. Therefore, no work is done by gravity.<br /><br />## Step5<br />For question 11, work is the product of force and distance. The force due to gravity on the 50.0 kg crate is \(50.0 \times 9.8 = 490\) N. The distance is 2.0 m, so the work is \(490 \times 2.0 = 980\) Joules. For the 25.0 kg crate, the force is \(25.0 \times 9.8 = 245\) N and the distance is 4.0 m, so the work is \(245 \times 4.0 = 980\) Joules. Both require the same amount of work.<br /><br />## Step6<br />For question 12, the work done in lifting the child is \(100 \times 1 = 100\) Joules. Carrying the child horizontally does not do any work against gravity. Therefore, the total work done is 100 J.<br /><br />## Step7<br />For question 13, work is the product of force and distance. Given that 2.0 J of work is done, and the force is \(0.18 \times 9.8 = 1.764\) N, the distance can be found using the formula \(distance = \frac{work}{force}\). Plugging in the values, \(distance = \frac{2.0}{1.764} = 1.13\) m.