Question

2. A block of 3 kg is sitting at rest on an incline of 10 degrees. A constant pushing force of 40 N is applied in the uphill direction for 1.3 seconds. The incline also provides a constant friction force of 20 N. Use the impulse-momentum approach to calculate how long the block will continue to slide before it comes to a stop on the incline.

Answer 1

Net force - F. F-40N may sina ingesa mg along the inclined for fost 1.3 second- Enet E-mgsino-fr = 40-3x9.8 xSino) -20 = 14.89N using Impulse- momentum relationship Fnet xt change in momentum Fret xt 3 m (22-9) = speed of block after 1:38 We initial speed of block =0 14.89 x 1:3 = 3x(2-0) Iv = 6.45 m/s by Work-Energy 2) if s be the distance travelled block during this time (t=138) then theorem - by Finet XS = 1 m (2²2) 14.09 X8 1x3?6.452-03} 8 = 62:40m het block-coveris & meter after removal force F-40N Net downward force Friet ng Sino + fr = 3x9.@xsinió +20 =25:11 N Now by Douk- Buergy ticerem- p=0 V=6.45 m/s s' K mgsing f ng ngoose

I m ( 22 - 262) =) z 2 2 501) - Fnet x8 (-ve sign es used in Lats because not force és opposite tie duflacement) -25.1188' = 4x3[03-6.45%) 1 x 3 x 6.452 s' = 2:49m So total distance Covered by block = 8+8! = 62.40 +7.49 64.89 m While distance travelled by block after Jeemoval 18'=2.49m of force,