On planet Tehar, the free-fall acceleration is the same as that on Earth, but th

Question

On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 4.7 kg ball having a charge of 2.74 ?C is thrown upward at a speed of 17.9 m/s, and it hits the ground after an interval of 3.85 s. The acceleration of gravity is 9.8 m/s^2.

What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.

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Explanation / Answer

charge is 2.74 C acc due to grvity = 9.8m/s^2 if we consider potentially energy =0 at initial point initially energy of ball = (mv^2)/2 = 4.7*(17.9^2)/2 finally kinetic energy will be zero so potential energy of ball at highest point = mgh+ potential energy due to electric field =752.9635 ball comes in 3.85 sec , so it reaches at its highest point at 3.85/2 = 1.925sec final velocity at highest point =v =0 equation of motion h = ut + a (t^2)/2 v^2 = u^2 + 2 a h =0 , a = -(u^2)/2h h = ut - (ut)^2/4h 4*h^2 - h *4* 17.9 * 1.925 - (17.9*1.925)^2 =0 h= 41.59 so mgh = 4.7*9.8*41.59 = 1915.6354 so potential energy due to electric field = 752.9635 - 1915.6354 = -1162.6719 so potential = -1162.6719/q = -1162.6719/2.74 = -424.332810219 V difference = 0 - - 424.332810219 = 424.332810219 V = .424 kV

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