1. A block of mass 15 kg slides down a frictionless ramp making an angle of 30o with the horizontal, as shown below.

 

 

 

 

 

 

 

 

 

 

a.  Redraw the block and draw in all the forces on the block (free body diagram)

 

 

 

 

 

 

 

 

 

b.  Choosing a coordinate system with one axis parallel to the ramp, write out Newton’s 2nd Law (F=ma) in both the x and y directions.

 

 

 

 

 

 

 

 

 

c.  Calculate the normal force, FN, (including the correct sign).

 

 

 

 

 

 

 

d.  Calculate the acceleration of the block (including the correct sign).

 

 

 

 

 

e.  Calculate how far the block will travel down the ramp from rest in a time of 1.5 s.


2.  A block of mass 5 kg is attached to a spring with spring constant, k = 200 N/m, and hangs down so that it rests on a table as shown below. This causes the spring to stretch by 10 cm from its equilibrium length. Assume the static coefficient of friction between the table and block is 0.6.  An external force, F, is applied in the positive x-direction, as shown.

 

 

 

 

 

 

 

 

 

 

a.  Redraw the block and draw in all the forces on the block (free body diagram)

 

 

 

 

 

 

 

 

 

 

b.  Calculate the force on the block due to the stretched spring.

 

 

 

 

 

 

 

c.  Using the coordinate given in the figure above, write out Newton’s 2nd Law (F=ma) in both the x and y directions.

 

 

 

 

 

 

 

 

 

 

d.  Calculate maximum force, F, that can be applied to the block such that the block will continue to remain at rest.

 

 

 


PHY203

Exam #2

Crib Sheet

Chapters 4,5

 

 

(Note: Use 9.81 m/s2 for g, the acceleration due to gravity.)

 

 

Constant acceleration:

 

xf = xo + vot + (1/2)at2

 

vf = vo + at

 

Vf2 = vo2 + 2a(xf - xo)

 

 

 

(Note: Bold letters indicate vectors below.)

 

 

F = ma

 

spring force: F = -kDx , where k is the spring constant

 

weight: W = mg

 

friction force: 

      kinetic        fk = mkFN , where FN is the normal force and mk is                                                                  the kinetic frictional coefficient

      static          fs < msFN,      fsmax = msFN

 

 

uniform circular motion

 

centripetal force:  F = mv2/r

 

centripetal acceleration:  a = v2/r

 

period:  T = 2pr/v