1. A 1500kg car takes a 50m radius unbanked curve at 15m/s. What is the size of

Question

1. A 1500kg car takes a 50m radius unbanked curve at 15m/s. What is the size of the friction force on the car?

2. A new car is tested on a 200m diameter track. If the car speeds up at a steady 1.5m/s, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?

3. A popular pastime is to see who can push an object closest to the edge of a table without it going off. You push the 100g object and release it 2m from the table edge. Unfortunately you push a little too hard. The object slides across the edge, falls 1m to the floor, and lands 30cm from the edge of the table. If the coefficent of kinetic friction is .5, whatwas the object's speed as you released it?

Explanation / Answer

1)

The Centripetal force required for the turn is provided by the frictional force only

so, Centripetal force = mv^2/r = 1500*15^2/50 N= 6750 N

This is also the friction force

2)

The tangential acceleration is actually 1.5 m/s^2, tangential acceleration is correct term for (tip acc) as tangential velocity is with tip speed.

centripetal acceleration = v^2 / radius

centripetal acc must equal 1.5 m/s^2

1.5 * radius = v^2

1.5 * 100 = v^2

v = sqrt(150) = 12.247 m/s

t = 12.247/1.5 = 8.165 secs (ANSWER)

3)

Use D = v*t + 0.5at^2       for vertical drop to get the time

1.0 = 0.5*g*t^2

t = sqrt(1*2/9.8) = 0.4517 sec

Use this equation again for horizontal motion to get velocity

0.3 = v*0.4517 + 0

v = 0.664 m/s

This is the velocity it leaves the table

As you rightly said now we can find the initial velocity

On the table the force resisting motion is friction = 0.5 * Normal Force

Normal Force = 0.1 * g

So Friction = 0.5*0.1*9.8 = 0.49 N

Using F=ma

0.49 = 0.1*a

a = 4.9 m/s^2

This is its decelaration along the table

Now use v^2 = u^2 + 2as

0.664^2 = u^2 - 2*4.9*2.0

u = sqrt (0.664^2 + 2*4.9*2.0) = 4.47 m/s {ANSWER}

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