Question

A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in this lift, assuming the object was lifted at constant velocity? (Include the sign of the value in your answer.) J (b) What total force (in N) was exerted on the man's teeth during the lift? (Enter the magnitude.) N

Answer 1

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}$

(a) Calculating the work done:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

$\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})$

$\text { Work }=-15.03 \mathrm{J}$

The work done by the man is -15.03J.

(b) Calculating the force:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is $\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

$\mathrm{F}=13 \times 9.8$

F = 127.4N

The force exerted on the object is 127.4N.