A1.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 29.0 ◦ with the horizontal, as shown in the figure. the acceleration of gravity is 9.81 m/s 2 . if the coefficient of kinetic friction between the block and the wall is 0.20, find a) the work done by the force on the block. answer in units of j. b) the work done by gravity on the block. c) the magnitude of the normal force between the block and the wall. answer in units of n

a) 35.08 J

b) -19.62 J

c) 17.53 N

Explanation:

A figure is attached in order to understand better the question. In addition, we have the following data:

is the mass of the block

is the distance the block is pushed at constant velocity

is the angle at wich the force is applied

is the acceleration due gravity

is the coefficient of kinetic friction between the block and the wall  

Now, if we draw a free body diagram of this situation we will have the following:

Where is the Normal force and is the horizontal component of the applied force

Then: (1)

Where is the vertical component of the applied force , is the Friction force and is the force due gravity (the weight of the block)

Then: (2)

Finding from (2):

(3)

(4)

(5)

When the applied force and the direction of motion form an angle the expression to calculate the Work is:  

(6)

(7)

(8) This is the work done by

In this case the work equation is:

(9)

Where:

is the gravity force on the block

is the angle between the gravity force and the direction of motion

(10)

(11) Note the work is negative because the applied force is in the opposite direction of motion

From (1) we know the Normal force is:

(1)

Since , we have:

(12)

(13) This is the Normal force

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answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6