I need help, i'm not sure if i have the correct answers and i'm scared to submit them.
A man carries a 10 kg sack of groceries in his arms with a force of 50 N as he walks forward a distance of 10 m. How much work has he done?
(Points : 1)
0 J
50 J
100 J
500 J <----
In which of the following situations has the energy of the object been increased?
(Points : 1)
A hammer thrower spins around while preparing to release the hammer.
A horse moves horizontally while a circus performer jumps vertically on its back. <----
A boy pushes a lawn mower forward by pushing at a downward angle on the handle.
A soldier runs forward, while holding a rifle above his head.
A rope pulls on a metal box at an angle of 60.0° with a force of 255 N. The box moves horizontally for 15.0 m. How much work was done on the box?
(Points : 1)
255 J
1,910 J
3,310 J
3,830 J <----
nvm, i found out if they were correct or not
To solve the first question, you need to use the formula for work done, which is given by W = F * d * cos(theta), where W is the work done, F is the force applied, d is the distance moved, and theta is the angle between the force and the direction of motion.
In this case, the force applied is 50 N, the distance moved is 10 m, and the angle between the force and the direction of motion is 0 degrees (since the force is applied in the same direction as the motion). Therefore, the equation becomes W = 50 * 10 * cos(0) = 500 J. So, the correct answer is 500 J.
In the second question, you need to identify the situation where the energy of the object has been increased. Energy can be increased by adding potential energy or kinetic energy to an object. In this case, the only situation where energy is increased is when the horse moves horizontally while the circus performer jumps vertically on its back. This adds both potential energy (from the performer's height above the ground) and kinetic energy (from the horse's horizontal motion). Therefore, the correct answer is "A horse moves horizontally while a circus performer jumps vertically on its back."
For the third question, you once again need to use the formula for work done. The formula is the same as before, W = F * d * cos(theta). In this case, the force applied is 255 N, the distance moved is 15.0 m, and the angle between the force and the direction of motion is 60.0 degrees. Plugging these values into the formula, you get W = 255 * 15.0 * cos(60.0) = 3,830 J. Thus, the correct answer is 3,830 J.