Ask question

Ask question

All categories

3 years ago

10

A 50.0 kg student climbs 5.00m up a rope at a constant speed. If the student's power output is 200.0 W, how long does it take th

e student to climb the rope ? How much work does the student do ?

1 answer:

hammer [34]3 years ago

3 0

Answer:

The time taken by the student to climb is 12.25 seconds and work done is 2450 J.

Explanation:

Given that,

Mass of the student, m = 50 kg

The student climbs to a height of 5 meters at a constant speed.

The student's power output is 200.0 W, P = 200 W

The power of an object is given by work done divided by time taken. So,

$P=\dfrac{W}{t}$

(b) W is work done,

$W=mgh\\\\W=50\times 9.8\times 5\\\\W=2450\ J$

(b)

$t=\dfrac{W}{P}\\\\t=\dfrac{2450}{200}\\\\t=12.25\ s$

So, the time taken by the student to climb is 12.25 seconds and work done is 2450 J.

You might be interested in

2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential

Citrus2011 [14]

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick = 7.9kg

Height of building = 22m

Unknown:

Potential energy of the brick = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

Insert the given parameters and solve;

P.E = 7.9 x 9.8 x 22 = 1703.24J

6 0

3 years ago

Which countries that had sea ice along their coasts in September 1986 were bordered by open water in September 2017?

mrs_skeptik [129]

Answer:

Norway, Sweden, Finland and Iceland

Explanation:

Sea ice is a frozen seawater that floats on the ocean surface. It is formed between the Antarctic and Arctic hemisphere. It disappears in summer but not completely. The countries that experienced sea ice in 1986 were eight (8) in number but the countries bordered by open water were in September 2017 were Norway, Iceland, Finland and Russia.

5 0

3 years ago

About 10-15% of all galaxies are which shape?

myrzilka [38]

The answer is; Irregular.

4 0

4 years ago

What force is required to accelerate a body with a mass of 15kilograms at a rate of

Paladinen [302]

The force required is

(15 kg) x (the acceleration, in m/s²) newtons.

4 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago