A shopper in a supermarket pushes a cart with a force of 38 N directed at an angle of 25 ◦ downward from the horizontal.Find the work done by the shopper on the cart as the shopper moves along a...
A shopper in a supermarket pushes a cart with a force of 38 N directed at an angle of 25
◦ downward from the horizontal.
Find the work done by the shopper on the cart as the shopper moves along a 48.6 m length of aisle. Answer in J.
Work done is equal to the the straight distance (s) moved by the multiplied by the magnitude of the force (f) in the direction of movement.
In the given question the cart moves by a distance of 48.6 m in horizontal direction.
The force applied (F) is 38 N in a direction making angle (A) of 25 degrees with the horizontal.
We can resolve this force in two equivalent components, one along the direction of movement of cart, and the other at right angles to it
The force along horizontal is given by:
f = F*(Cos A) = 38*(Cos25) = 38*0.9063
As the cart moves only in horizontal direction and there is no other movement in any other direction, the total work done (w) is that due to movement of cart.
Therefore: w = s*f = 48.6*(38*0.9063)
= 1673.7548 J
Answer: Work done by the shopper on the cart is 1673.7548 J
The force of pushing the cart is 38N with a direction 25 degree below horizontal.
We assume that the aisle on which the cart is moving is horizontal:
The force along the direction of the cart = 38N cosine (angle25 deg) only.
Work done is the product of the force along the direction of movement cart and the distance the cart moved along the aisle = (38N cos 25)*48.6m
=1830.5534 Joule of work is done.
If the aisle itself is 25 degree inclined to the horizontal and the cart needs a constant force of 38 deg below horizontal horizontal, then full component of the 38N force is along the aisle . therefore work done by 38N is 38N*48.6 cosine ( angle zer) = 38N*48.4m*1 = 1846.8 Nm =1846.8 Joules of work.