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SOVA2 [1]
3 years ago
5

A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

and the two cars skid forward 2.4 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact. What was the speed of the sports car at impact?
Physics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction (\mu)=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

v^2-u^2=2as

acceleration=\mu g=0.8\times 9.8=7.84 m/s^2

s=2.4 m

0-(u)^2=2\times (-7.84)\times 2.4

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

920\times v=(920+2300)\times 6.13  (conserving momentum)

v=21.45 m/s

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3 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
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v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

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Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

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but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

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In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

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Answer:

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