1) Draw a free body diagram for each of the following objects:
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1 Forces - Pracice 1) Draw a free body diagram for each of he following objecs: a) An apple (a res) siing on a able. b) A mass of 10 kg hanging from a single rope (a res, no swinging) c) A man falling off a cliff, halfway down in fligh (wih no air resisance) d) An objec ha was pushed and is now sliding across he floor wih no fricion. e) Same as #4 bu now wih a fricion force ha is 20% of is weigh free-body diagrams: a) b) c) Normal Tension Weigh Weigh Weigh d) Same as a since no acceleraion e) Same as a bu wih a vecor in he horizonal direcion labeled Fricion which opposes he moion f) Sae Newon s 3 Laws. You should know hese! 2) A man has a mass of 75 kg. a) Deermine his weigh on he Earh Weigh = Force = mass x g = 750 N b) Deermine his mass in space, far from any graviaional objecs. Mass is he same no maer where you are c) Skech a free-body diagram for he man as he i) is sanding on he floor a res See a in he free body diagrams above ii) is in free fall afer he jumped off a cliff See c in FBD above
2 The man is now sanding on he ground and holds a 25 kg objec over his head d) Skech a free body diagram for he man Normal Weigh by Earh Force on man by 25kg objec e) Deermine he value and direcion of he normal force on him by he floor. Normal = Weigh of Man + Weigh of objec = 750N + 250N = 1000 N 3) A ne force of 10 Newons is applied on an objec in he horizonal direcion. I goes from 2 m/s o 8 m/s in 10 seconds. Deermine he mass of he objec. F = ma mass = Force / acceleraion Force = 10 Newons Solving for acceleraion: a = Δv Δ = v f = 8 m s 2 m s 10s = 0.6 m s 2 m = Force acceleraion = 10N =16.67kg 0.6 m s 2 4) If a 60 kg mass undergoes an acceleraion of 5 m/s 2 due o a force F, wha will he acceleraion be if: a) he same force F is applied o a mass of 90 kg Mah soluion, solve for original Force and plug ino new equaion F = ma = 60kg 5 m s 2 = 300N now plug in a = F m = 300N 90kg = 3.33 m s 2 Alernae Soluion! Since a = Force/mass, here is an inverse relaionship beween acceleraion and mass Answer mass goes up by 3:2 raio, hen acceleraion has o go down by 2:3 raio, or, wo hirds he original value. 2/3 of 5 is 3.33
3 b) a force of one enh of F is applied o a mass of 6 kg a = F m = 30N 6kg = 5 m s 2 Noe: a = F / m They boh wen down same % c) a force of ½ F is applied o a mass of 6 kg a = F m = 150N 6kg = 25 m s 2 5) A meal box has a mass of 10 kg and is siing a res on he floor. a) Skech a free body diagram for he box when i is a res See free body diagram a on firs page The meal box is now pushed on by a Space Monkey (who has 100% fricional shoes). The objec ravels a disance of 400 cm and hen leaves he Space Monkey s hand wih a speed of 4 m/s (assume i acceleraed consanly). The floor has some fricion. The Space Monkey s force on he box was 30 Newons and o he righ. b) Deermine he acceleraion of he box while he Space Monkey pushed i. Fricion Opposes (lef), he Force by he Monkey is greaer, o he righ. Ne Force = 30N minus Fricion Force Iniial velociy = 0 Final Velociy = 4 m/s Disance = 4 m (400 cm) v f 2 = v i 2 + 2ax Solving for acceleraion v 2 2 f 2x = a = (4 m s ) m = 2 m s 2 c) Deermine he ne force on he box while he Space Monkey pushed i. ΣF = ma =10kg 2 m s 2 = 20N e) Give a reason why he answer o par c differen han he value of he force applied. There was fricion acing on he box. The value of he fricion mus have been 10N
4 f) Draw a free body diagram for he objec while he Space Monkey is pushing i. See par b of his problem. For he Y direcion, see he FBD problems, a g) Deermine he value and direcion of he fricion force. 10 Newons, opposing he moion 30N-20N=10N Quesions (h) and (i) are from he ime he objec leaves he Space Monkey s hand unil i comes o res a disance away. Assume he fricion force remains he same as (g). h) Deermine how far he box will slide before i comes o res Use Work-Energy heorem All he kineic energy is used up by he force of fricion. Ne Work = ΔKE Work = F x KE = 1 2 mv 2 Work = Fx = 1 2 mv 2 x = 1 mv F = 10kg(4 m 2 s )2 10N i) How much ime elapses unil i comes o res? = 8m Remember, he fricion force is -10N (opposing moion), using F=ma we ge a = -1m/s 2 Use equaion for acceleraion a = v f Δ Solve for ime Δ = v f a = 0 4 m s 1 m s 2 = 4s 8) You push agains a wall on a fricionless surface, and weigh 600 N. Afer your hand leaves he wall, you end up moving o he lef 20 meers in 5 seconds. a) Skech a free-body diagram for you while you are in conac wih he fall. Three vecors: Forces in Y cancel ou (weigh and normal), however, he wall pushes back on you, in he opposie direcion ha you push on he wall b) If he conac ime was 0.1 seconds, deermine he force you applied on he wall. This one is ricky. Hidden in he quesion is he final velociy of he push by on wall on you. How? You wen 20 meers in 5 seconds on fricionless surface and i didn say you come o res. I simply gives daa o deermine your speed speed = Δx Δ = 20m 5s = 4 m s When you sared he push, your velociy was zero. So now you can deermine he acceleraion, and wih ha, he Force since F=ma. Recall, your mass is 60 kg since weigh is 600N (W=mg) F = ma = m v f = 60kg 4 m s 0.1s = 2400N
5 c) Wha ype of force acceleraed you? Which law applies? The normal force by he wall on you is wha acceleraed you. This is he acion-reacion pair of you pushing on he wall d) If he conac ime was 0.2 seconds, find he average force you applied on he wall. Here, you are geing a change in velociy over a longer ime period, so i will be a smaller force: F = ma = m v f = 60kg 4 m s 0.2s =1200N e) How would he answers o (b) and (d) be differen (do he mah) if someone who weighed 500 N performed he same experimens wih he same ime and disance values. The free-body diagram would be differen, because you weigh less. I would require a smaller force o ge o he velociy of 4 m/s. Using 50-kg insead of 60-kg 2000 N for b, and 1000 N for d f) If he force you applied (from b) was consan, skech a velociy vs. ime graph for you during he momen you sared o push unil 1 second laer. Use he whole graph. Find he slope of your graph. Wha does i give you? Wha if you muliplied he slope by your mass? v 4 m/s 1 s ime Slope gives you acceleraion. Muliplying he acceleraion by your mass gives you he NET FORCE on you! Noice afer 1 second he slope is ZERO, which means he ne force on you MUST BE zero as well.
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