Alistair places a 6.40 kg block on an inclined plane that is 23° above the horizontal. The block starts from rest, and in 2.20 seconds, travels 1.40 m. What is the force of friction on the block? O 19.5 N 3.70 N 20.8 N 24.5 N
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- Given are the three forces F, = (2.27 + 2.2)N, F, = (2.27 - 2.2)N, and F3 = (-2.27 + 2.2) N. Which of the following vectors is net 1 Select one: Oa. R= (-2.2î+2.2) N Ob. R= (2.2î+2.2) N Oc. Ř= (6.6 7+ - 2.2)N Od. R= (2.2î+6.6 ) N Oe. Ř= (2.2 î+2.2) NDetermine the magnitude of the x- component of force Q (in N) if P=96.4 N, Q = 55.2 N , 0= 55.53° and a = 49.46°.Round off only on the final answer expressed in 3 decimal places. Instead of units, indicate the direction of the x-component. Use R if to the right, L if to the left, U if upward or D if downward. y PA 250 gram block on a 30 deg inclined plane is attached by an inextensible string going over a frictionless pulley to a 500 gram hanging weight. The static and kinetic friction coefficients are .25 and .123. While moving, what is the tension ii the connecting string? O 174530 dynes O 121000 dynes O 262400 dynes O 7430 dynes O 42930 dynes
- A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle 0 below the horizontal, and the box remains a rest. The box rests on a horizontal surface that has some friction with the box. The normal force on the box due to the floor is equal to: O W + P cos 0. O W + P sin 0. OW-P sin 0. O W + P. O W. W 0 P14. A 25.0 kg girl goes down a slide at an amusement park, reaching the bottom. The slide is 10.0m long and the top end is 3.00m above the ground, bottom is 0.30m above the ground. The coefficient of friction between the girl and the slide is 0.150. What velocity will she exit, at the bottom of the slide. Draw a FBD, explain what you are doing h₁Calculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that force as the only unknown. The forces are positive if in tension, negative if in compression. 3.0 m BC= Answers: BE= A 3.0 m EF= G i i i B 3.0 m 3.0 m F 1.9 m E 3.0 m 20.0 KN KN KN kN D
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