Work is done only when there is a displacement of the object on which a force is applied in the direction of the force.
The answer to your question is meters.
The force acting on the ball are unbalanced. Reactionary momentum force (that originated as a result of the swing of the bat) is the most powerful.
Yes friction is acting on the ball. In course of journey it would slow the ball down and make it trace a parabolic path rather than straight path as intended by hitter.
Explanation:
As the hitter hits the ball, momentum of the bat due to swing (mass of the bat*velocity provided by the batsman swinging action of bat) gets transferred on the ball on its impact with the bat.
Since ball’s mass is quite small as compared to the bat, the velocity of the ball increases by the same factor by which the ball’s mass is lower than the bat’s mass. This velocity causes forward motion of the ball (of course in the direction of bat’s motion, here the batsman intends to send the ball straight away hence the ball would move straight).
Various forces on ball is-
- Reactionary momentum force -bat’s force (most powerful force)
- The frictional force of the air (opposing the motion of the ball through the air)
- Gravity force (pulling the ball down to the Earth)
As a combined effect of these force when all the force remains unbalanced, the ball moves away in the straight path under the impact of bats momentum which was most powerful of all.
Frictional force and Gravity force continue acting on the ball. While frictional forces decrease the ball velocity through the air, gravity force pulls it down thus deflecting its direction. Under the combined impact of declining bats momentum, friction force and gravity force, the ball traces a parabolic path (in accordance with the first law of motion from Newton)
Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
