Mathematics, 26.09.2019 18:30 nyameneak

Afactory worker pushes a 30.0 kg crate a distance of 4.3 m along a level floor at constant velocity by pushing downward at an angle of 30∘ below the horizontal. the coefficient of kinetic friction between the crate and floor is 0.25.

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Answer from: victoriabrooks5766

a) the worker must apply 73.5 Newtons

b) work done = 330.75 Joules

c) work done = -330.75 Joules

Work done by normal force = 0 Joules

Work done by gravity = 0 Joules

e) total work done = 0 joules

Step-by-step explanation:

The complete question:

"A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?"

distance, s = 4.3

mass, m = 30

Coefficient of friction, $\mu_k=0.25$

The worker here must apply a force, $F_{worker}$ equal to kinetic frictional force F_k .

We know $F_k=\mu_k mg$

We know g = 9.8, thus plugging in we get:

$F_{worker}=F_k=\mu_k mg=0.25*30*9.8=73.5$

Thus, the worker must apply 73.5 Newtons

Work is given by the formula:

$W=F_{worker} s Cos\phi$

F and s are in the same direction, so angle ( $\phi$ ) is 0

Now we find work done:

$W=F_{worker} s Cos\phi\\W=73.5*4.5*Cos(0)\\W=73.5*4.5\\W=330.75$

Hence, work done = 330.75 Joules

We use same formula but remember that the angle is 180 degrees since frictional force is acting against it (opposite).

Thus we have:

$W=F_{worker} s Cos\phi\\W=73.5*4.5* Cos(180)\\W=-330.75$

Hence, work done = -330.75 Joules

Work on by normal force would be perpendicular, so angle would be 90 degrees. THus

$W_n=F_{worker} s Cos (90)\\W_n=0$

Work by gravity would be opposite of that direction. That would be -90 degrees. Thus

$W_g=F_{worker} s Cos(-90)\\W_g=0$

Thus

Work done by normal force = 0 Joules

Work done by gravity = 0 Joules

Net work done on crate will be sum of all individual work done.

$W_{net}=W{worker}+W_k+W_n+W_g$

Plugging in the values, we have:

$W_{net}=W{worker}+W_k+W_n+W_g\\W_{net}=330.75-330.75+0+0=0$

Hence

total work done = 0 joules

Note: this is 0 because F_worker and F_k are working opposite to each other

Answer from: Quest

what is the original number? you forgot to add it.

Answer from: Quest

step-by-step explanation:

the graph shows 30% of students walk. 80 students were questioned, so the number that said they walk is

30% × 80 = 0.30 × 80 = 24

Agroup of 80 students was asked to share how they get to school most of the time. the following circ

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Afactory worker pushes a 30.0 kg crate a distance of 4.3 m along a level floor at constant velocity...

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