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yuradex [85]
2 years ago
7

How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.

)
A) 100 J B) 200 J C) 300 J D) 400 J
Physics
1 answer:
alisha [4.7K]2 years ago
4 0

Answer:

B) 200 [J]

Explanation:

In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]

F = m*g

where:

m = mass = 5 [kg]

g = gravity acceleration = 10 [m/s^2]

F = 5*10 = 50 [N]

w = F*d

where:

F = force = 50 [N]

d = 4 [m]

w = 50*4 = 200 [J]

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ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces
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Answer:

The magnitudes of the second force is   Z = 129.9 N

The magnitudes of the  resultant force is   R = 256.047 N

Explanation:

From the question we are told that  

    The force is  F = 187 \ lb

     The angle made with second force \theta_o = 73 ^o 36' =  73 + \frac{36}{60}  =  73.6^o

     The angle between the resultant force and the first force \theta _1  = 29 ^o 1 ' = 29 + \frac{1}{60}  = 29.0167^o

For us to solve problem we are going to assume that

     The magnitude of the second force is  Z N

     The magnitude of the resultant force is R N

According to Sine rule

                \frac{F}{sin (\theta _o - \theta_1 }  = \frac{Z}{\theta _1}

Substituting values

             \frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}  

             267.82 =\frac{Z}{0.4851}  

              Z = 129.9 N

According to cosine rule

       R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }

Substituting values

     R = \sqrt{187^2 + 129.9 ^2  + 2 (187 ) (129.9) cos (73.6)}

     R = 256.047 N

 

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