When the volleyball rises in the air, what does the force of gravity cause the volleyball to do

Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

Sidana [21]

The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons

5 0

3 years ago

A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

$F_s = \mu N$

where $\mu$ is the coefficient of friction and $N$ is the normal force.

When the student pulls on the block with force $F_p$ at an angle $\theta$ , the normal force on the block becomes

$N = Mg- F_psin(\theta)$

and hence the frictional force becomes

$F_s = \mu (Mg- F_psin(\theta))$ .

Now, as we increase $\theta$ , $sin(\theta)$ increases which as a result decreases the normal force $Mg- F_psin(\theta)$ , which also means the frictional force decreases; Hence choice B stands true.