A child pushes a toy box 5.0 m along the floor with a force of 7.0 N directed downward
at an angle of 30.0° to the horizontal. (a) How much work does the child do? (b) Would
you expect more or less work to be done if the child pulled upward at the same angle?
the force in the direction of motion is ... 7.0 N * cos(30º)
(a) w = 7.0 N * cos(30º) * 5.0 m
(b) less ... pulling upward reduces the frictional force
please, i need the answer ASAP
To determine how much work the child does in pushing the toy box and whether more or less work would be done if the child pulled upward, we need to calculate the work done in both cases.
Work (W) is defined as the product of the force applied (F) and the displacement (d) in the direction of the force:
W = F * d * cos(θ)
where θ is the angle between the direction of the force and the direction of displacement.
(a) In this case, the force applied by the child is 7.0 N directed downward at an angle of 30.0° to the horizontal, and the displacement of the toy box is 5.0 m. Therefore, plugging in these values into the work formula, we have:
W = 7.0 N * 5.0 m * cos(30.0°)
To calculate the cosine of 30.0°, we can use a calculator or lookup table, which gives us cos(30.0°) = 0.866.
W = 7.0 N * 5.0 m * 0.866
W ≈ 30.3 J (rounded to one decimal place)
So, the child does approximately 30.3 Joules of work in pushing the toy box.
(b) If the child pulled upward at the same angle, the force would be directed upward at an angle of 30.0° to the horizontal. The displacement would remain the same (5.0 m). Following the same steps as before, the work done can be calculated as:
W = 7.0 N * 5.0 m * cos(30.0°)
Using the same cosine value as before, we find:
W = 7.0 N * 5.0 m * 0.866
W ≈ 30.3 J (rounded to one decimal place)
Therefore, if the child pulled upward at the same angle, the amount of work done would be approximately the same (30.3 J) as when pushing the toy box downward.