a womans lifts a 300 newton child a distance of 1.5 meters and carries her forward for 6.5meters.
how much work does the woman do in lifting the child?
b.how much work does the child do?
Women
W=f x d
W= 6.5 n x 1.5 m = 9.75 J
Child
0J
To calculate the work done, we can use the formula:
Work = Force × Distance
a. To calculate the work done by the woman in lifting the child, we need to multiply the force exerted by the woman with the distance lifted.
Given:
Force = 300 Newton
Distance = 1.5 meters
Work = 300 N × 1.5 m = 450 Joules
Therefore, the woman does 450 Joules of work in lifting the child.
b. To calculate the work done by the child, we need to consider that the child is being carried, so the work done by the child can be considered negligible or nearly zero. Since the child is not exerting any force, no work is done by the child.
To calculate the work done, we need to use the formula:
Work = Force * Distance * Cos(θ)
Where:
- Force is the amount of force applied
- Distance is the distance over which the force is applied
- Cos(θ) is the angle between the direction of the force and the direction of the displacement
a) To calculate the work done by the woman in lifting the child, we have the following values:
- Force (F) = 300 Newtons (given in the question)
- Distance (d) = 1.5 meters (given in the question)
- Angle (θ) = 0 degrees (since the force is applied vertically)
Plugging these values into the formula:
Work = 300 N * 1.5 m * Cos(0°)
Since Cos(0°) = 1, the equation simplifies to:
Work = 300 N * 1.5 m * 1 = 450 Joules
Therefore, the woman does 450 Joules of work in lifting the child.
b) The child is not doing any work in this scenario because the child is being carried without exerting any force or energy. The child is simply being lifted and carried by the woman.