Elastic strings and springs Mixed exercise 3

Transcription

1 Elastic strings and springs Mixed exercise ( )T cosθ mg By Hooke s law 5mgx T 6a a sinθ a + x () () () a 4 5mg Ifcos θ, T from () 5 8 5mg 5mgx so, from () 8 6a a x 4 If cos θ, then sinθ 5 5 a sinθ from () a a+ 5 4 which is true. So, cosθ. 5 b Work done on particle overall gain in energy P.E. gain E.P.E. loss PM ( a+ x) cosθ E.P.E. loss initial E.P.E. final E.P.E. a 4 5mg a a a 4a 5mg 4a 6 a 9 4a 5mga P.E. gain mg 4mga 5mga So, work done mga (6 5) mga Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

2 Let l be the natural length of the spring. Let λ be the modulus of the spring. ( ) T W by Hooke s law, λa T l λa W λ W i.e. l a l Using conservation of energy, P.E. loss ofw E.P.E. gain of spring a λx W + x l a Wx so, W + x a a + ax x 0 x ax a 0 ( x a)( x+ a) x a or a maximum compression is a Substitute for l λ from above. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

3 a x λ 0.5 by Hooke s law,5 0.5 λ 0N b (final) (initial) Using conservation of energy K.E. gain E.P.E. loss E.P.E. loss initial E.P.E. final E.P.E. v So, v ( ).J v. ( ).9 m s (s.f.) Initial E.P.E (( ) + ( ) ) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.

4 4 Triangle ABP is a,4,5 triangle, so angle APB is a right angle. 4 cos θ and sinθ 5 5 ( ) T sinθ + T cosθ mg 4 T + T mg 5 5 4T + T 5mg () ( ) T cosθ T sinθ 4 T T T T () Substituting from () into (): 6 T + T 5mg 5T 5mg T mg 5 From Hooke s law, λx λ( a l) T l l mg a λ 5 l mg a + 5λ l mg+ 5λ a 5λ l 5λa l mg+ 5λ Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 5 a ( ) R mg Friction mg 5 Work done against friction overall loss in energy 5 mg a E.P.E. loss K.E. gain a 5mg mv a ag 5ag V 5 4 5ag ag ag(5 ) V V ag 0 b Friction will be the same. Assume string is still slack when ball comes to rest. Work done against friction K.E. loss V 4V mgd m m ag gd a d 50 As d is less than a, the assumption that the string is still slack is valid. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 6 a ( ) R mg ( ) µ R T by Hooke s law, µ mg T mg l mg T l mg µ mg µ b Work done against friction overall loss in energy E.P.E. loss K.E. gain mgl mgl mv l V gl V gl gl gl V gl c String is now slack. Work done against friction K.E. loss mg d m gl d l Total distance travelled is l. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 7 a extension of AP ( x ) 0.cosθ 0.5 extension of BP ( x ) 0.sinθ 0.05 x x ratio is b ր 0.cosθ sinθ cosθ 4sinθ ( )along PB: T 5gcosθ ( ւ)along PA: T 5gsinθ T so, T cosθ sinθ λx 0.5 cosθ 0.05 λx sinθ x cosθ x sinθ x sinθ i.e. x cosθ Using the answer to part a: 4cosθ sinθ 4sinθ cosθ sin θ(4sinθ ) cos θ(4cosθ ) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 8 a ( ր perpendicular to string) 9.8cosθ gsinθ tanθ θ tan 8.4 b ( ) Tsinθ 9.8 T x x.08 The extension is. m ( s.f.). c Least force will be perpendicular to the string ( ր) F gsinθ g 0 g The least force is 9. N ( s.f.). Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 9 a By conservation of energy, K.E.gain+ E.P.E.gain P.E.loss mg x mv + mg 4a 4 a BP 5 a (, 4,5 triangle) So, x 4a mg 6a mv + 4mga 4 a V 4ga 8ga + V 4ga V ga b mg 4a x 4 a: T 4 a mg Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 Challenge We define the variables for the problem as in the following diagram; a Now, applying conservation of elastic potential energy and gravitational potential energy, we find that: λ mgd ( d l) l λ mgld + mgl d d l + d l λ ( d l dl) Substituting mgl k and rearranging, we see that: λ d ( l+ k) d+ l 0 ( ( l+ k) ( l k) l ) d ± + d l+ k ± k + ( ) lk But we know that d must be larger than l, else the string wouldn t be taut when the maximum depth was reached, so we should take the positive square root, giving the result. b i Suppose the jumper had an initial downwards velocity, v. Then they would have an initial kinetic energy mv in the downwards direction, in addition to the initial GPE of mgd. So the distance the jumper falls increases. ii If we included air resistance, the frictional force would do work on the jumper as they fell. Then the energy balance is GPE + EPE + Work done by friction 0. This results in a reduced GPE, decreasing the maximum distance the jumper falls. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0