A shopper pushes a grocery cart for a distance 18 m at constant speed on level ground, against a 25 N frictional force. He pushes in a direction 24° below the horizontal.
What is the work done on the cart by the shopper, in joules? Magnitude of the force, in newtons, that the shopper exerts on the cart? What is the total work done on the cart, in joules?
Fx - Fk = M*a
Fex*Cos24 - 25 = M*0 = 0
Fex*Cos24 = 25
Fex = 27.4 N. = Force exerted
Work = Fx*d = 27.4*Cos24 * 18 = 450.6 Joules
To find the work done on the cart by the shopper, we can use the formula: work = force * distance * cos(theta)
Where:
force = magnitude of the force exerted on the cart by the shopper
distance = distance traveled with the cart
theta = angle between the force exerted by the shopper and the direction of motion of the cart
Given:
distance = 18 m
force of friction = 25 N
theta = 24° below the horizontal
First, let's find the magnitude of the force exerted by the shopper on the cart.
We can use the force of friction as the opposing force.
The force exerted on the cart by the shopper can be represented as the vertical component of the force.
So, the force exerted by the shopper can be calculated as:
force = force of friction / sin(theta)
force = 25 N / sin(24°)
Using a calculator, we find that the force exerted by the shopper on the cart is approximately 57.27 N (rounded to two decimal places).
Next, let's calculate the work done on the cart by the shopper.
work = force * distance * cos(theta)
work = 57.27 N * 18 m * cos(24°)
Using a calculator, we find that the work done on the cart by the shopper is approximately 921.37 J (rounded to two decimal places).
Finally, let's calculate the total work done on the cart. Since the force of friction is acting in the opposite direction of motion, the total work done on the cart will be the negative value of the work done by the shopper.
total work done = -921.37 J
Therefore, the work done on the cart by the shopper is 921.37 J (rounded to two decimal places). The magnitude of the force exerted by the shopper on the cart is approximately 57.27 N (rounded to two decimal places). The total work done on the cart is -921.37 J (rounded to two decimal places).
To find the work done on the cart by the shopper, we can use the equation:
Work = Force * Distance * cos(theta)
where:
- Work is the work done on the cart by the shopper (in joules),
- Force is the magnitude of the force exerted by the shopper on the cart (in newtons),
- Distance is the distance the shopper pushes the cart (in meters), and
- theta is the angle between the direction of the force exerted by the shopper and the direction of the displacement of the cart.
In this case, the distance the shopper pushes the cart is given as 18 m. The frictional force against the cart is 25 N. The angle between the direction of the force exerted by the shopper and the direction of motion of the cart is 24° below the horizontal.
First, let's find the force exerted by the shopper on the cart:
The vertical component of the force is given by:
Force_vertical = Force * sin(theta)
Force_vertical = 25 N * sin(24°)
Force_vertical ≈ 10.17 N
The horizontal component of the force is given by:
Force_horizontal = Force * cos(theta)
Force_horizontal = 25 N * cos(24°)
Force_horizontal ≈ 22.69 N
To find the magnitude of the force exerted by the shopper on the cart, use the Pythagorean theorem:
Magnitude of Force = sqrt(Force_vertical^2 + Force_horizontal^2)
Magnitude of Force ≈ sqrt(10.17 N^2 + 22.69 N^2)
Magnitude of Force ≈ sqrt(103.48 N^2)
Magnitude of Force ≈ 10.17 N
Now, let's find the work done on the cart:
Work = Force * Distance * cos(theta)
Work = 10.17 N * 18 m * cos(24°)
Work ≈ 176.92 J
So, the work done on the cart by the shopper is approximately 176.92 joules. The magnitude of the force exerted by the shopper on the cart is approximately 10.17 newtons.