All categories

3 years ago

A weightlifter is bench-pressing 710 N. He raises the weight 0.65m above his chest. How much work does he do with this lift?

Physics

2 answers:

stepan [7]3 years ago

6 0

The work done is 461.5 J.

Explanation:

According to Kinetics, the work done is calculated as the product of force which has done work, and the displacement it has imposed. Hence,

$\text {Work done}=F \times s$

In the problem, the given data, F = 710 N and s = 0.65 m

By applying the given values to the work done equation, we get as follows,

$\text { Work done } = 710 \times 0.65 = 461.5 \mathrm{J}$

The unit is same as the energy, work done defines when 1 N force acts in a 1 m distance along the force’s direction.

Assoli18 [71]3 years ago

4 0

The work done is 6958 J

Explanation:

The work done by the man to raise the weight is the equal to the increase in gravitational potential energy of the weight, so we can write:

$W=mg \Delta h$

where

(mg) is the weight, with m being the mass and g the acceleration of gravity

$\Delta h$ is the change in height of the weight

In this problem, we have

mg = 710 N (weight)

$\Delta h = 0.65 m$

Therefore, the work done is

$W=(710)(9.8)=6,958 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

You might be interested in

Why is raw sewage a major pollutant in some countries but not in developed countries?

Damm [24]

Answer:because developed cities are more equipped to deal with raw sewage than less developed cities

Explanation:

3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes: