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Luba_88 [7]
3 years ago
14

A 75.0-kg painter climbs a ladder that is 2.75 m long leaning against a vertical wall. The ladder makes a 30.0° angle with the w

all. (a) How much work does is done by gravity on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?
Physics
2 answers:
Schach [20]3 years ago
6 0
<h3><u>Answer;</u></h3>

a) 178.6125 Joules

b) Work done does not depend on constant speed or acceleration up the ladder

<h3><u>Explanation</u>;</h3>

Work done =Fd times the cosine of angle between;

where F is force d is distance m is the mass a is acceleration

.F=ma. In this case a=g=9.81 m/s^2 and cos 30=0.866

Therefore;  

W= 75 × 9.81 × 2.75 × .866  

   = 178.6125 Joules

b. The work done by the painter does not depend on whether the painter climbs at constant speed or accelerates up the ladder.

W=3376.945 Joules

Mazyrski [523]3 years ago
4 0

Answer:

Work is defined as the movement caused by a force. In the case of someone climbing something, the person is doing force against gravity, so the work needed to lift an object of mass M by a height H is equal to W = MgH where h is the gravity acceleration.

In this case, M = 75kg and we can obtain the height by the Pythagorean theorem, where the ladder is the hypotenuse of a triangle.

If we want to only know the displacement in the y-axis, then we need to compute the adjacent cathetus, this is:

H = 2.75m*cos(30°) = 2.38m

and we know that g= 9.8m/s^2

so we can compute the work and obtain:

W = 2.38m*9.8m/s^2*75kg = 1739,3J

b) work does not depend on the velocity, the fact that he is ascending means that there is a force applied on him that fights against the gravity. Non the less, if there was a force bigger applied to him in the y-direction (which may accelerate him, the previous calculation is thinking that the force done is equal to the gravitational force) the work can be bigger. The work calculated before is the minimal work needed to make him ascend the height H.

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022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

v_f = 0 + (90)(0.33)

v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

6 0
3 years ago
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