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Problem With Physics ...physics Gurus Please Help by BjDreamchaser(m): 6:49pm On Feb 09, 2017 |
please I need help and detailed explanations to solving these questions.. 1. A 300g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2. 2. A horizontal force of 200N is required to cause a 15kg block to slide up a 20° incline with an acceleration of 25cm/s2. Find (a) the friction force on the block and (b) the coefficient of friction. 3. A lawn tennis ball of mass m and speed v strikes a wall perpendicularly and rebounds with undiminished speed. If the time of collision is t, what is the average force exerted by the ball on the wall? 4. A block of mass 2kg is pulled along a smooth horizontal surface by a horizontal force P. What is the value of the normal force exerted on the block by the surface, and determine the force P required to give the block horizontal velocity of 4ms-1 in 2s |
Re: Problem With Physics ...physics Gurus Please Help by BiafranBushBoy: 7:36pm On Feb 09, 2017 |
How much will you pay? Let me tackle it once and for all.... |
Re: Problem With Physics ...physics Gurus Please Help by Nobody: 7:47pm On Feb 09, 2017 |
Firstly, I'm not a guru Secondly, I may be wrong. I'll attempt 3 and 4 For question 3, according Newton's second law. F = change in momentum/time From the question, neither mass nor velocity changes... Hence F = Mv - Mv/t F = 0/t F= 0... i.e no force was applied for question 4, although this is quite confusing; a. the force exerted by the surface should be the Normal Reaction. Hence, R = mg R = 2 * 10(provided g=10m/s); R= 20N b. Using Newtons second law F = M(v - u)/t.... Though it doesn't state the block was at rest. Assuming it was F = 2(4 - 0)/2; F= 4N |
Re: Problem With Physics ...physics Gurus Please Help by Nobody: 3:38pm On Feb 17, 2017 |
BjDreamchaser:Soln M¹=300/1000= 0.3kg M²=900/1000= 0.9kg a=0.700m/s² g=10m/s² (a) Tension = mg + ma (upward acceleration) =0.3*10 + 0.3*0.700 =3 + 0.21 =3.21N Tension in second string => 0.9*10 + 0.9*0.700 9+0.63 =9.63 Since the second mass hangs on M¹ the total force/tension in string one will be 3.21N + 9.63 (the extra force pulling M¹) =12.84N (b) Tension with downward acceleration = mg - ma 0.3*10 - 0.3*0.700 3 - 0.21 =2.79N Second string 0.9*10 - 0.9*0.700 =9 - 0.63 =8.37N Total Tension in First string with downward acceleration. . =8.37 + 2.79 =11.08N |
Re: Problem With Physics ...physics Gurus Please Help by BjDreamchaser(m): 3:55pm On Feb 17, 2017 |
Optimist1998: 100% right. Thank you boss... |
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