The wheel of a car has a radius of 0.350 m. the engine of the car applies a torque of 295n⋅m to this wheel , which does not slip against the road surface . since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a counter-torque. moreover, the car has a constant velocity , so this counter-torque balances the applied torque . what is the magnitude of the static frictional force?

842.85 N

Explanation:

Given

Engine of car applies a Torque of 295 N.m

radius of wheel r=0.350 m

since the car is moving with constant velocity therefore net torque on it must be zero

Let friction force force applies a counter torque given by

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.