Slide14.JPG

Slide15.JPG
Slide16.JPG
Slide17.JPG


Transcript

Example 9.5 The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in Fig. 9.9. How much force does the table exert on the ball to bring it to rest? How much force does the table exert on the ball to bring it to rest? From the graph, we note that At Time = 0 s, Initial velocity = u = 20 cm/s At Time = 10 s, Final velocity = v = 0 cm/s We need to find Force Finding Acceleration first Since Velocity – Time graph is a straight line The acceleration is constant. Acceleration = (πΉπ‘–π‘›π‘Žπ‘™ π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ βˆ’ πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)/π‘‡π‘–π‘šπ‘’ a = (𝑣 βˆ’ 𝑒)/𝑑 = (0 βˆ’ 20)/10 = (βˆ’20)/10 = βˆ’2 cm/s2 Converting to m/s2 = (βˆ’2)/100 m/s2 = βˆ’0.02 m/s2 Now, Mass of the body = 20 g = 20/1000 kg = 0.02 kg We know that Force = Mass Γ— Acceleration Force = 0.02 Γ— 0.02 = (βˆ’2)/100 Γ— 2/100 1 m = 100 cm 1 cm = 1/100 m 1 cm/s2 = 1/100 m/s2 = (βˆ’4)/1000 = βˆ’0.004 N The Negative sign of the force means that force was applied in opposite direction of motion

Ask a doubt
Maninder Singh's photo - Co-founder, Teachoo

Made by

Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo