Question

question 1 A block of mass m is pulled at constant velocity along a rough horizontal floor by an applied force � as shown. What is the magnitude of frictional force? Write your answer in terms of m, g, T, or q.

question 2 An 8000-N car is traveling at 12 m/s along a horizontal road when the brakes are applied. The car skids to a stop in 4.0 s. How much kinetic energy does the car lose in this time? How much work is done by the friction of the brakes?

question 3 A 2-kg object is moving at 3 m/s. A 4-N force is applied for 1s in the direction of motion and then removed. Calculate the work done by this force.

question 4 The standard 1-kg mass is attached to a compressed spring and the spring is released. If the mass initially has an acceleration of 5.6 m/s2 , what is the magnitude of the force of the spring?

please explain and show all work because i don't know anything

Answer 1

0 +2)!- Consider the diagram as shown below! N a=o, 'Tsino T L'ex (عے گر) Tcoso + ng where, fx= fictional force. Normal force N= Normal bla block and table. Apply Newton's and how on the Given system as shown below: System EFx = mxa. Fu = = mxo 'Thoso - fk = I fx= T cos of And =)

Q-(2) - consider the Given datang shown below:- weight sof Car ( 1) = 8000 N. Initial velocity of Car (4) 12 m/s brake is applied & Car stofs, So, finial velocity (6)=0 m/s. mass of Car (m) = w 1kg) g g= accelsation due to grarity = 9,82 m/s Now, detsmine the loss in Kie (ok.e)" by using following formula as shown below:

(KE) = (finial IC. 2) - (Initial (ie) 2 mx wa 2 Clc.E) > 1 شده - values m, in above Eqn as shoven below! okie) = 12 x wcG-42) g wo U, put value of og u,2g in En(1) as shown below: ५००० (1.E) = 80oor (10) ² (12) 2) 9.81 - 576000 -58715.596(J) 9.81 010-4)=– 5.871x10 (J) | Anx

[when the Ensgy is lost in system, then, indicates that Enisgyis loxt] (ie, t-ve) sign is obtaine in answes, it means that Enggyis lost . = -5.871x10 J And Now, upe, worl energy theorem on Given system. (K.C)i + bl.s)i + (We) = (k~of+ (p.5/5 (p.5) 2 = (pit) = 0) (Car is moring on horizontal Surface) wf = work done by friction

-> (W) = (K.Elf - VC. 2); - Ock.s). Inf= o(K-2)= o(K-4)= - 5.871x10 J Ana W Q-B)! Consider the Giren denta as shown below:- mass C cm) 2kg. Initial velocity (u) = 3 m/s. applied () = (t) = 1 (8) Force - 2-N. Now, The accelsation of object is, F = mxa. Ta= 1/ m 2 m/s Z

Now defrmine the distance travelled (s)" by the object by application of Civen force is as shown below:- 2 (s) = ((ext) + 1 (9) x (H) I Ceyn of motion) (s) = (3x1) + 1x 125 x (L) ² - s = (3 + 1) = 4 m Now, detsimine the work love force (w) as shown below:- W=Fx sx coso (~)

put the value of F, s, &020, in Eqa (2) as shown below: (W) = 4x4x Coso (W) = 16 J . And mass 8-14).- Corisider the Given dataas shown below:- of block (m) = 1 kg accefsation in block (a)= 5-6 m/s magnitude of force in Afring (45)=2 draw the free body diagram as shown below:-

AN a) (Fs) ng x-direction as Newton and raw in Apply Shonen below :- & EF = (mxa) (+ Fs) = (mxa) Fs = (2x 5.6) = 5.6N/. Are

Therefore, 8-10% magnitute of frictional force is (fx) = Tease And 8-(2)! (K.E) loss in Caris, 5.871x10 J. Eks) - work done bu by friction, Ws) - 5.871x10) Ang. Q-13%. work done by force is, Tw= 16 J Ang. Two magnitude of force in spring is, 9-14:- Fra sig lang