In what unit do we measure the work done by a force

Physics

1 answer:

Gala2k [10]3 years ago

7 0

Explanation:

the work done by a force is measured in Newton(N) in SI unit and in CGS unit it is measured in dyne

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You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

Mrac [35]

The general formula to calculate the work is:

$W=Fd \cos \theta$

where F is the force, d is the displacement of the couch, and $\theta$ is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.

(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so $\theta=0^{\circ}$ and $\cos \theta=1$ , therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

$W=Fd=(220.0 N)(3.9 m)=858 J$

(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore $\theta=180^{\circ}$ and $\cos \theta=-1$ . Therefore, we must include a negative sign when we calculate the work done by the frictional force:

$W=-Fd=-(144.0 N)(3.9 m)=-561.6 J$

(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore $\theta=90^{\circ}$ and $\cos \theta=0$ : this means

$W=0$ .

(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

$F=220 N-144 N=76 N$

And the net force is in the same direction of the displacement, so $\theta=0^{\circ}$ and $\cos \theta=1$ and the work done is

$W=Fd=(76 N)(3.9 m)=296.4 J$

4 0

3 years ago

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

solong [7]

Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s