Web6m The 300 N crate slides down the curved path from A to B in the vertical plane as shown in the figure. If the crate has a velocity of 1.2 m/s down the inclined at point A and 8 m/s at B. Find the work done against friction during the motion (g = 10 m/s) (A) 818 Nm (B) 987 Nm (C) 1200 Nm - 9m (D) none of these Solution Verified by Toppr Web27 Aug 2024 · The 300 N crate slides down the inclined curved path from A to B in the vertical plane as shown in the figure. If the crate has a velocity of 1.2 m/s down the …
A 1 kg block situated on a rough inclined plane is …
Web26 Mar 2016 · Notice that the crate’s mass does not affect its acceleration because mass cancels out of the final equation. Sample question. A plastic crate slips down a 19-degree ramp with a coefficient of kinetic friction of 0.10. What is its acceleration as it slides? The correct answer is 2.3 m/s 2. You can use the equation WebA 20-N crate starting at rest slides down a rough 5.0-m long ramp, inclined at 25° with the horizontal. 20 J of energy is lost to friction. What will be the speed of the crate at the bottom ... b. 5 300 N c. 6 500 N d. 7 700 N 55. A pole vaulter clears 6.00 m. With what speed does he strike the mat in the landing area? a. 2.70 m/s b. 5.40 m/s onrealm login church
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WebExpert Answer. A 3.00-kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.0° as shown in the figure. The crate starts from rest at the top and experiences a constant friction force of magnitude 5.00 N. The crate continues to move a short distance on the horizontal floor after it leaves the ramp, and then ... WebThe lateral bars are specially designed to allow air to feed the fire, resulting in a clean and efficient burn. Our Fireplace grates are sized to the British standard classification, ranging … Webthe 50-kg, uniform crate slides down the inclined surface with an initial speed 10 m/s. if the coefficient of kinetic friction between the crate and the inclined surface is 0.15, determine the distance s the crate has moved when it momentarily stops. (Assume tipping over does not happen) Expert's answer l=v^2/ (2a) l = v2/(2a) onreceivelocation bdlocation location