Physics question about a bulldog skateboarding through a dip in the road: In the
ID: 1475258 • Letter: P
Question
Physics question about a bulldog skateboarding through a dip in the road:
In the three figures labeled A, B, and C below are shown three situations in which a bulldog on a skateboard traverses a dip in the ground. (The dips are all the same.) In A, the bulldog enters the dip traveling at a non-0 kinetic energy (KE). In B, when he leaves the dip, he has 0 KE. In C, he begins in the dip (at the position indicated) at a KE of 0. Since he is riding a skateboard, friction may be ignored. Take g 10 N/kg. th--KE > 0 here KE = 0 here KE 0 here A. In the figures below are shown graphs of the KE, gravitational PE, and total energy as a function of position. In situation 1, the total energy is negative; in situation 2, the total energy is 0; and in situation 3, the total energy is positive. Match which situation goes with which graph and put the letter of the situation to the left of the corresponding graph.Explanation / Answer
A)
For situation1: The P.E is zero at the start of the motion with K.E > 0 and hence when the bulldog moves down the dip it increases its P.E in negative direction(w.r.t 0 P.E mark) and there is increase in the kinetic enegry so, the total enegry is constant
hence graph 3 holds.
Situation 2 : the final K.E is zero and cosidering energy conservation the total energy must be zero as the complete P.E is converted into K.E.
hence graph 2 is correct
Situation 3: Analysing the situation as explained above graph 1 is correct.
B)
Applying energy conservation from bottom of the well to the point when it comes out; it can be concluded that the final P.E and K.E are zero
so, P.Ei + K.Ei = P.Ef + K.Ef
-mgh + 1/2 mvi2 = 0+0
h = 1.9m
C)
the K.E of a body is due to its motion and is therefore always positive whereas the P.E is due to the positon of the body w.r.t some reference point so it can be negative or positive. A negative T.E implies that the body is having more negative P.E or zero K.E at some point under observation. At the lowest point of the dip the P.E is dominating than K.E and hence giving a negative T.E
D)
given total energy at the lowest point
so, T.E = K.Ei + P.Ei
200 = K.Ei +(- P.Ei ) ..................(i)
Applying energy conservation between bottom point to final point
K.Ei + P.Ei = K.Ef + P.Ef
K.Ei + P.Ei = 0+0
K.Ei = -P.Ei .............(ii)
from (i) and (ii) K.E = 100J