What is ihe minimum acceleration that the car in the figure must have at the top

Question

What is ihe minimum acceleration that the car in the figure must have at the top of the track if it is to remain in contact with the track? a. 4.9 m/s^2, downward b. 4.9 m/s^2, upward c. 9.8 m/s^2, upward d. 9.8 m/s^2, downward e. 19.6 m/s^2, upward 19. A 10 kg block is pushed against a vertical wall by a horizontal force of 100 N. The coefficient of static friction mu_s, between the block and the wall is 0.60. and the coefficient of kinetic friction mu_k is 0.40. Which of the following is true if the block is initially at rest? a. the tolal force exerted on the block by the wall is horizontally directed b. the block slides down the wall with an acceleration of 3.8 m/s^2 c. the block will slide down the wall because the force of static friction can be no larger than 60 N d. the block will remain at rest because the coefficient of static friction is larger than the coefficient of kinetic c. the net external force is zero 20. A block of mass m slides with constant velocity v down a plane inclined at theta with the horizontal. During the lime interval Deltat, what is the magnitude of the energy dissipated by friction? a. mgvDeltat sin theta b. mgvDeltat cos theta c. mgvDeltat tan theta d. (1/2)V^3 Deltat e. The answer cannot be determined without knowing the coefficient of friction 21. On an icy winter day, the coefficient of friction between the tires of the car and a roadway might be reduced to one-half of its value on a dry day. As a result, the maximum speed at which a curve or radius R can be safely ncgolialed is a. the same as on a dry day b. reduccd to 71% of its value on a dry day c. reduced to 50% of its value on a dry day d. reduccd to 37% of its value on a dry day

Explanation / Answer

At top of the track there is no normal force, Weight of the Car will be acting Downwards, This would cause the Centripetal Force.

So,
m*a = m*g
a = g = 9.8 m/s^2
This is directed away from the center, i,e Upwards.
Ans :-   9.8 m/s^2, Upwards

Energy Dissipated by Friction,
Friction Force, F = uk*N
F = uk*m*g*cos()

Distance, = v*t

Energy = Force * distance
Energy = uk*m*g*cos()*v*t

So we can see, Energy cannot be detrmined without Friction.

Initially,
mv^2/r = Friction Force

Now, Coefficient of Friction is reduced to half, so Friction Force is also reduced to half.
mVnew^2/r = Friction Force/2
mVnew^2/r = (mv^2/r)/2
Vnew = 1/sqrt(2) * v
Vnew = 0.707 v

So it is reduced to 71% of it's value on a dry day.

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