Physics, 19.03.2020 08:09 Magree5321

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).

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Answer from: richardwalker8ourhg2

(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.

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Answer from: Quest

a=d/t

a="triangle"v/t

explanation:

a = acceleration

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A 15-kg block is dragged over a rough, horizontal surface by an applied force of [01] N acting at 20...