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Lorico [155]
3 years ago
6

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

ge friction force of 4.0 ✕ 103 N impedes the motion, find the speed of the car at the bottom of the driveway.
Physics
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m=2.1\times 10^3 kg

Initial velocity,u=0

Distance,s=5.9 m

\theta=19^{\circ}

Average friction force,f=4.0\times 10^3 N

We have to find the speed of the car at the bottom of the driveway.

Net force,F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3

Where g=9.8 m/s^2

Acceleration,a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}

v=\sqrt{2as}

v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}

v=3.9 m/s

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A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

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The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
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