Question

it has a speed When the 7-kg box reaches point VA = 13 m/s. (Figure 1) Part A Determine the normal force the box exerts on the surface when it reaches point B. Neglect friction and the size of the box. Express your answer to three significant figures and include the appropriate units. Figure < 1 of 1 I Å O2 ? N N = = Value N Submit Previous Answers Request Answer x1/2 + yl/2 = 3 X Incorrect; Try Again; 5 attempts remaining Provide Feedback Ne:

Answer 1

Given data :- m: 7 kg ,VA= 10mis drace the free body system as shown in diagram the figure of the belo- am mg cosotmus? from the given equation, Fn't ry za differenti at the above equation, dy I dat 250 217

I dr. – I de 22* at point B, yen at point Biyon from ego U .' 0 - 5 lope Edy - 7 any termo-7 0 - =) 03-75 125 from Ho energy conservation at A and B. mgß's Inhuma ' = VA? 112 28 288.81 : 8.61 m

find a position at sng. from the equation dat Jy =a Int In=3 lm- g = 2.25 m Apply to energy conservational highest point and point B. Єo - Ед 36 = a+ ra ở UR? = eg (4-9) Vaz 2x9.81% [ 8.61- 2,25) B. 124:18 De = 11:17 mis reaction at point B. The normal Mamupa R t mg cosa

Rz7x 110 + ing cose R - from the given equation- 469-21) (4 min) and arz analna ल* on 2.2 dy an2 - EP = 0.429. 7-2.25 The radius of Cereature of point B iso le al2/

0:449 R- 6.36 m. put the valle in ego . N=78 11-192 +7 49:81X COS (545) 6:36 N= 137.32 + 48:56 PE 185.00 80.

m-20 F-500p , S= 0.5 on Consider the figure shown below from the question as ' ܢܕܘܘܟ find the soon k= 500 Pim @ - toim ) al = 36.9699 Apply the work energy conservation. Fxs = { muz tdks2 500 x Cos 36.969 x 0.5 - In Doxu 2 5 X 50px 0:52 200 I lov2 + 6215 U2=13.75 V= 3.7 mis