Question

A 15-kg block is dragged over a rough, horizontal surface by an applied force of [01]____________________ N acting at 20.0 degrees above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300. Find the work done by a. the applied force (J). b. the normal force (J). c. the force of gravity (J). d. the force of friction (J).

Answer 1

Explanation:

Lets us suppose that the applied force is 70 N.

In the vertical direction, the normal force, mg, and the vertical component of F.

From the FBD in attachment we have,

a) work done by applied force = Fcosθ×d= 70cos20°×5.0=328.89N

also, b) N + Fsin20°= mg

⇒N = mg - Fsin20° = 15.0×9.8 - 70×sin20° = 123.06 N

Normal force does no work because there is no displacement in that direction. Similar is the case with force of gravity.

d) The frictional force f = μN = 0.3×123.06 = 36.92 N

So the force done by friction force f_w = -fd = -36.92×5 = -184.59 J

the increase in the internal energy of the block-surface system due to friction is 184.59 J.

Answer 2

Answer:

(a). The work done by the applied force is 328.89 J.

(b). The work done by the normal force is zero.

(c). The work done by the force of gravity is zero.

(d). The work done by the force of friction is 184.5 J.

Explanation:

Given that,

Force = 70 N

Mass of block = 15 kg

Angle = 20°

Displaced = 5.00 m

Coefficient of kinetic friction = 0.300

(a). We need to calculate the work done by the applied force

Using formula of work done

$W=Fd\cos\theta$

Put the value into the formula

$W=70\times5.00\cos20$

$W=328.89\ J$

(b). We need to calculate the work done by the normal force

The normal force is perpendicular to the displacement of the motion,

So it does not work.

(c). We need to calculate the work done by the force of gravity

The gravitational force is perpendicular to the displacement,

Or the displacement in the vertical direction is zero.so the work done is zero.

(d). We need to calculate the normal force

Using balance equation

$N+f\sin\theta=mg$

$N=mg-f\sin\theta$

Put the value into the formula

$N=15\times9.8-70\sin20$

$N=123.0\ N$

We need to calculate the force of friction

Using formula of friction

$F=\mu N$

Put the value into the formula

$F=0.3\times123.0$

$F=36.9\ N$

We need to calculate the work done by the force of friction

Using formula of work done

$W = F\cdot d$

Put the value into the formula

$W=36.9\times5.00$

$W=184.5\ J$

Hence, (a). The work done by the applied force is 328.89 J.

(b). The work done by the normal force is zero.

(c). The work done by the force of gravity is zero.

(d). The work done by the force of friction is 184.5 J.