Question

friction
up until this point, its been convenient to ignore one of th
most troublesome forces in everyday life... a force that ha
plication in transportation, machinery, and all parts of mec
spend tremendous amounts of effort and money each da
force, friction, is a force that opposes motion.
4.24 q: a projectile launched at an angle of 45° above t
s through the air. compared to the projectil
with no air friction, the actual trajectory of the
friction is
(a) lower and shorter
(b) lower and longer
(c) higher and shorter
(d) higher and longer
4.25 q: a box is pushed toward the right across a classroon
force of friction on the box is directed toward the
(a) left
(b) right
(c) ceiling
(d) floor
4.28 q: the diagram below shows a 4.0 - kilogram object ac
10 meters per second² on a rough horizontal surfac
acceleration = 10. m/s²
frictional
force = f₁
m = 4.0 kg
applied
force = 50. n

Explanation:

4.24 Q:

When there is air friction (air resistance), it acts opposite to the direction of the projectile's motion. In the vertical direction, air friction reduces the upward velocity, so the maximum height reached is lower. In the horizontal direction, air friction decelerates the projectile, reducing its horizontal range. So the trajectory is lower and shorter compared to the no - air - friction case.

Friction is a force that opposes relative motion. The box is moving to the right, so the force of friction acts in the opposite direction (to the left) to oppose this motion.

Step1: Apply Newton's second law

Newton's second law is \(F_{net}=ma\). The net force \(F_{net}\) is the applied force \(F_{applied}\) minus the frictional force \(F_f\), i.e., \(F_{net}=F_{applied}-F_f\).

Step2: Substitute the values

We know \(m = 4.0\space kg\), \(a=10\space m/s^{2}\), and \(F_{applied}=50\space N\). From \(F_{net}=ma\), we have \(F_{applied}-F_f=ma\). Substituting the values: \(50 - F_f=4\times10\).

Step3: Solve for \(F_f\)

First, calculate \(4\times10 = 40\). Then, rewrite the equation as \(F_f=50 - 40\).

Answer:

A. lower and shorter

4.25 Q: