A block of ice that weighs 510 newtons is pushed up a 6 m long incline. The incline is also 3 m tall.
a) Neglecting friction, how much force is needed to push it up the incline?
b) How much work is required to push the ice block up the incline compared to lifting the block vertically 3 m?
a. neglicting friction...what a dream. force must be 510(3/6)
b. work will be the same, neglecting friction.
To answer these questions, we need to consider the principles of work and energy.
a) Neglecting friction, the force needed to push the ice block up the incline can be determined using the concept of mechanical advantage. The force required can be calculated using the formula:
Force = Weight x (Height of incline / Length of incline)
In this case, the weight of the ice block is given as 510 newtons. The height of the incline is given as 3 m and the length of the incline is given as 6 m. Plugging these values into the formula:
Force = 510 N x (3 m / 6 m) = 255 N
Therefore, neglecting friction, a force of 255 Newtons is needed to push the ice block up the incline.
b) The work done to push the ice block up the incline can be calculated using the formula:
Work = Force x Distance
In this case, we have already determined the force required to be 255 Newtons. The distance traveled up the incline is given as 6 m. Plugging these values into the formula:
Work = 255 N x 6 m = 1530 Joules
To compare this with lifting the block vertically 3 m, we can use the formula for work done when lifting an object against gravity:
Work = Force x Distance
The force required to lift the block vertically is equal to the weight of the block, which is 510 Newtons. The distance lifted vertically is given as 3 m. Plugging these values into the formula:
Work = 510 N x 3 m = 1530 Joules
Thus, the work required to push the ice block up the incline is the same as the work required to lift the block vertically 3 m, which is 1530 Joules.