Question

The driver of a car of mass M which is moving along a straight road with initial speed v₀ sees a deer in her headlights, and reacts quickly, lifting her foot of the gas and applying the brake pedal with maximum force. The anti-lock brakes cause the largest possible static friction force to be applied on the tires by the road, which continue to roll so the car does not skid. The coefficient of static friction between the tires and the road is μ_s. The car comes to a stop at a distance d from where the driver first applied the brakes, and the startled deer slowly walks away, unharmed.

Part 1 Over the distance the car moves while the brakes are applied, the work done by the static friction force acting on the car is (positive, negative, or zero)?

Part 2 The direction of the static friction force on the car tires due to the road is, relative to the direction in which the car is moving (up, right, left, backward, down, or forward)?

Part 3 Choose all answers which could equal the work done by the static friction over the distance the car moves while the brakes are applied.

- − ½ M v₀²

- 0

- ½ M v₀² − μ_s M g d

- − μ_s M g d

- − μ_s M g

- ½ M v₀²

- μ_s M g d

- μ_s M g

Part 4 If the mass of the car is 1,073 kg, its initial speed is 19.6 m/s, and the distance it moves while braking and coming to a stop is 31.7 m, what is the coefficient of static friction between the tires and the road?

Answer 1

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Lhx10hp 21 ox all we'll do this quis hon principle; by using energy conservath Cod Total initial Kinetic ener rgy to Work done by frictional force =0 Now, frictional force, Fj = Ms Mg & rintto a

Date Page & Kinetic energy Cinitially ] = 1 1 Muot; thus work done by frictional energy I Mu ? 2 + Work done 7 Work done = - LMvo? Also Force Bo displacement work done Now, frictional force & the direction in enact opposite to car are 180 or of pach other, Work done done Us My x dx cos180°) This Work Mg Mgd by prictional force is negative direction of frictional direct is meg opposite As the explaned above, relative to s backward car force je + AS cuplaned above, frictional work done Muo ² & M = 1073 Vo = 19.6mls, d = 21.7m 2 1 x 107 1 x 1073 x 19.6 = Msx 1073 xq 81*31.7 2 us = 0.618