1) A force of 157 N is applied to a box 25o above the horizontal. If it is applied over a distance of 14.5 m how much work is done?


2) Jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. How fast will it be falling right before it hits the ground?


3) Snookie is riding in her little red wagon (total mass of 215 kg) at a constant speed of 9.5 m/s. All of a sudden a magenta lemur (19.7 kg) appears in her lap. How fast is the Snook-meister traveling now?


4) Police are investigating an accident. They know that Tom Brady was driving 20 m/s before being hit by Jay Z head on. Tom Brady’s car has a mass of 1100 kg and Jay Z’s has a mass of 1475 kg. They also know that the two cars stuck together and were traveling 7 m/s in the same direction as Jay Z was driving. The speed limit was 25 m/s, was Jay Z speeding?


5) A ball is being swung in a horizontal circle of radius .23 m at a rate of 7.8 m/s. If the ball has a mass of 0.63 kg, find the force of tension on the string.

Respuesta :

#1

Work done is given by

[tex]W = F.d cos\theta[/tex]

here we know that

F = 157 N

d = 14.5

[tex]\theta = 25^0[/tex]

now from the above formula

[tex]W = (157)(14.5)cos25 = 2063.2 J[/tex]

#2

By energy conservation

Initial total potential energy of egg = final total kinetic energy of egg

[tex]mgh = \frac{1}{2}mv^2[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.8)(8.2)}[/tex]

[tex]v = 12.7 m/s[/tex]

#3

By momentum conservation we know that

[tex]m_1v_1 = (m_1 + m_2)v[/tex]

[tex]215(9.5) = (215 + 19.7) v[/tex]

[tex] v = \frac{215 (9.5)}{215 + 19.7}[/tex]

[tex]v = 8.7 m/s[/tex]

#4

by momentum conservation we know that

[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]

let direction of velocity of Jay Z car motion is positive direction

now we will say

[tex]1100(-20) + 1475v = (1100 + 1475)(7)[/tex]

[tex]1475 v = 2575(7) + 1100(20)[/tex]

[tex]v = 27.14 m/s[/tex]

Yes Jay Z is overspeeding as his speed is more than speed limit

#5

Tension force in the string will be centripetal force

so here we can say

[tex]T = \frac{mv^2}{R}[/tex]

here we know that

m = 0.63 kg

v = 7.8 m/s

R = 0.23 m

[tex]T = \frac{0.63 (7.8)^2}{0.23}[/tex]

[tex]T = 166.6 N[/tex]

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