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zzz [600]
2 years ago
14

3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of

37° with the horizontal, as shown. If the cofficient of kinetic friction between the block and the wall is 0.30. Find the work done by the applied force the frictional force.​
Physics
1 answer:
Andrews [41]2 years ago
5 0

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

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A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v
hram777 [196]

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

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Get the net force:

Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

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Read 2 more answers
It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

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3 years ago
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