Question

You're driving your pickup truck around a curve that has a radius of 22 m.How fast can you drive around this curve before a steel toolbox slides on the steel bed of the truck?

Answer 1

The maximal speed around this curve before a steel toolbox slides on the steel bed of the truck : 14.832 m/s

Further explanation

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle

$\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}$

F = centripetal force, N

m = mass, Kg

v = linear velocity, m / s

r = radius, m

The speed that is in the direction of the circle is called linear velocity

Can be formulated:

$\displaysyle v = 2 \pi.r.f$

r = circle radius

f = rotation per second (RPS)

Pickup trucks that move around a curve will experience centripetal forces. So that the steel toolbox does not experience slides on the steel bed of the truck, the centripetal force experienced by the steel toolbox must be the same as its weight. If the centripetal force exceeds the weight of the steel toolbox, the steel toolbox will fall

centripetal force = weight

$\rm \dfrac{mv^2}{r}=mg$

then the maximal speed so that the steel toolbox doesn't fall :

$\rm v^2=r\times g\\\\v=\sqrt{r\times g}$

a curve has a radius of 22 m , then

$\rm v=\sqrt{22\times 10}\\\\v=\sqrt{220}\\\\v=\boxed{\bold{14.832\:m/s}}}$

Learn more

the average velocity

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resultant velocity

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velocity position

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Answer 2

The maximum speed of a truck around this curve before a steel toolbox slides on the steel bed of the truck is 14.832 m/sec.

What is centripetal force?

When a body is accelerated in a close curve a force towards its centers is acting known as centripetal force.

                                  $\rm{F_c=\frac{mv^{2} }{r}}$

where,    

F = centripetal force in N

m = mass in Kg

v = linear velocity in m / s

r = radius in m

According to conditions in order to avoid the slide of the box. The weight of the box must be greater than or equal to the centripetal force acting on the pickup truck.

$\rm{\frac{mv^{2} }{r}=mg}$

$\rm{v^{2} =rg}$

$\rm{v=\sqrt{rg}}$

given,

r = 22 m

g = 10m/s^2

$\rm{v=\sqrt{22\times10}}$

$\rm\\{v=\sqrt{220}}$

$\rm{v=14.832 m/sec}$

The maximum speed of a truck around this curve before a steel toolbox slides on the steel bed of the truck is 14.832 m/sec.

To learn more about the centripetal force refer to the link

https://brainly.in/question/23006